Let $G=SL\_n(\mathbb C)$ and $T$ be a maximal torus. Then the Grassmannians $Gr(r,n)$ and $G(n-r,n)$ are isomorphic. Now for the left action of the torus on each of them can we say that the GIT quotients $T \backslash \backslash G(r,n) (\mathcal L\_{n\omega\_r})$ and $T \backslash \backslash G(n-r,n) (\mathcal L\_{n\omega\_{n-r}})$ are isomorphic ? Here $\omega\_r$ and $\omega\_{n-r}$ are the fundamental weights associated to the simple roots $\alpha\_r$ and $\alpha\_{n-r}$ respectively and the quotient means the semistable quotients. 



Quotients of Grassmannians
I am writing this as an answer because the comments are already too long. In the following I am incredibly pedantic, because there seems to be endless possibility for confusion with the several simultaneous group schemes, group scheme elements, and group scheme automorphisms that are involved. 


Let $k$ be a field; later I will assume that $k$ is algebraically closed. Let $X$ be a projective $k$-scheme. Let $\mathcal{L}$ be an ample invertible sheaf on $X$. For instance, if $X$ is a Fano manifold (such as any projective homogeneous space), we might choose $\mathcal{L}$ to be the determinant of the tangent bundle. 


**Lemma 1.** The automorphism group $k$-scheme of $(X,\mathcal{L})$, $\text{Aut}\_k(X,\mathcal{L})$, is an affine group $k$-scheme. 


**Proof.** This is discussed many places, such as Subsection 2.1 of the following article.


A. J. de Jong and J. Starr   

Almost proper GIT-stacks and discriminant avoidance  

Doc. Math. 15 (2010), pp.957-972.


I am pretty sure that this is also in Grothendieck's Bourbaki seminar notes, "Fondements de la Géométrie Algébrique." **QED**


Let $T$ be a (geometrically) reductive group $k$-scheme. Let $$\rho:T\to \text{Aut}\_k(X,\mathcal{L})$$ be a morphism of group $k$-schemes. 


**Definition 2** The **semistable locus** of $\rho$ is the open subscheme of $X,$ $$X^{\text{ss}}\_\rho:=\cup\{ D\_{\mathcal{L}^{\otimes n}}(s) | n\in \mathbb{Z}\_{>0}, \ s \in H^0(X,\mathcal{L}^{\otimes n})^{\rho(T)}\}.$$ 


For every integer $n$, denote by $\gamma\_n$ the natural morphism of group $k$-schemes, 
$$\gamma\_n:\text{Aut}\_k(X,\mathcal{L})\to \text{Aut}\_k(X,\mathcal{L}^{\otimes n}).$$


**Lemma 3.** The open subscheme $X^{\text{ss}}\_\rho$ of $X$ depends only on the image of $\rho$. Also, for every positive integer $n$, $X^{\text{ss}}\_{\gamma\_n\circ \rho}$ equals $X^{\text{ss}}\_{\rho}$ as open subschemes of $X$.


**Proof.** The first statement is straightforward from the definition of the semistable locus, which only depends on the $\rho(T)$. For the second statement, for every $\rho(T)$-invariant section $s$, also $s^n$ is invariant under $\gamma\_n\circ\rho(T)$. Since $D(s)$ equals $D(s^n)$, it follows that the open subschemes are equal. **QED**


**Definition 4.** The **associated prerelation** $R^{\text{pre}}\_\rho$ is the minimal closed subscheme through which the following morphism factors, $$\Psi\_\rho:T\times\_{\text{Spec}\ k} X^{\text{ss}}\_{\rho} \to X^{\text{ss}}\_\rho \times\_{\text{Spec}\ k} X^{\text{ss}}\_\rho, \ \ (t,x)\mapsto (\rho\_t(x),x).$$ The **associated relation** $R\_\rho$ is the first iterate of this relation, i.e., the minimal closed subscheme through which the following morphism factors, $$(\text{pr}\_1\circ \text{pr}\_1,\text{pr}\_2\circ \text{pr}\_2):R^{\text{pre}}\_\rho \times\_{\text{pr}\_2,X^{\text{ss}}\_\rho,\text{pr}\_1} R^{\text{pre}}\_\rho \to X^{\text{ss}}\_\rho\times\_{\text{Spec}\ k} X^{\text{ss}}\_{\rho},$$ $$((x\_1,x),(x,x\_2))\mapsto (x\_1,x\_2).$$ 


**Fundamental Theorem of Geometric Invariant Theory (Projective Case).** [Theorem 1.10, p. 38, *Geometric Invariant Theory, 3rd ed.*, Mumford, Fogarty, Kirwan, Ergebnis. Math. 34, Spring-Verlag, Berlin, 1994]   
 Among all $G$-invariant $k$-morphisms from $X^{\text{ss}}\_\rho$ to $k$-schemes, there exists an initial such $k$-morphism, $$q:X^{\text{ss}}\_\rho \to Y,$$ i.e., $q$ is a categorical quotient. This morphism is a uniform categorical quotient, i.e., it is a categorical quotient after finitely presented, flat base change of $Y$. This morphism is affine. This morphism is universally submersive, i.e., for every finitely presented morphism $Y'\to Y$ and for every subset $U\subset Y'$, $U$ is open if and only if the inverse image of $U$ in $Y'\times\_{Y,q} X^{\text{ss}}\_\rho$ is open. As a closed subscheme of $X^{\text{ss}}\_\rho\times\_{\text{Spec}\ k} X^{\text{ss}}\_\rho$, the fiber product $R\_q:=X^{\text{ss}}\_\rho\times\_{q,Y,q} X^{\text{ss}}\_\rho$ contains $R\_\rho$, and they are equal as closed subsets. The scheme $Y$ is projective, and there exists an ample invertible sheaf whose pullback under $q$ with its natural linearization is isomorphic to a positive tensor power of $\mathcal{L}$ with its induced linearization. Finally, there exists an open subscheme $Y\_0$ of $Y$ whose inverse image equals the (properly) stable locus $X^{\text{s}}\_{\rho,0}$, and the restriction $q\_0:X^{\text{s}}\_{\rho,0}\to Y\_0$ is a uniform geometric quotient.


There is an important observation about the fiber product $R\_q$. Since $q$ is $G$-invariant, $R\_q$ contains $R^{\text{pre}}\_\rho$.
Since $R\_q$ is the fiber product of a morphism, this is an equivalence relation. Thus, $R\_q$ contains $R\_\rho$. The content of set-theoretic equality of $R\_q$ and $R\_\rho$ is the following. For every geometric pair of points, $$(x\_1,x\_2):\text{Spec}\ \kappa \to X^{\text{ss}}\_\rho\times\_{\text{Spec} k}X^{\text{ss}}\_\rho,$$ the images of $x\_1$ and $x\_2$ in $Y(\kappa)$ are equal if and only if the orbit closures of $x\_1$ and $x\_2$ in $X^{\text{ss}}\_\rho\times\_{\text{Spec}\ k}\text{Spec}\ \kappa$ have nonempty intersection. Moreover, every geometric fiber of $q$ contains a unique orbit that is closed in $X^{\text{ss}}\_\rho$. 


**Lemma 5.** A separated $k$-morphism $p:X^{\text{ss}}\to Z$ is invariant for the action $\rho$ if and only if the fiber product $R\_p:=X^{\text{ss}}\_\rho \times\_{q,Y,q}X^{\text{ss}}\_\rho$ contains $R\_\rho$ as closed subschemes of $X^{\text{ss}}\_\rho\times\_{\text{Spec}\ k}X^{\text{ss}}\_{\rho}.$


**Proof.** By definition, $p$ is $\rho$-invariant if and only if $R\_p$ contains the image of $\Psi\_\rho$. Since $p$ is separated, $R\_p$ is a closed subscheme. Thus, $p$ is $\rho$-invariant if and only if $R\_p$ contains $R^{\text{pre}}\_\rho$. In this case, since $R\_p$ is an equivalence relation, $R\_p$ contains the first iterate $R\_\rho$ of $R^{\text{pre}}\_\rho$. **QED**


Let $(X,\mathcal{L})$ and $(X',\mathcal{L}')$ be polarized projective $k$-schemes. Let $T$ and $T'$ be reductive group $k$-schemes. Let $$\rho:T\to \text{Aut}\_k(X,\mathcal{L}), \ \ \rho:T'\to \text{Aut}\_k(X',\mathcal{L}'),$$ be morphisms of group $k$-schemes. Denote the GIT quotients above by $$q:X^{\text{ss}}\_\rho\to Y, \ \ q':(X')^{\text{ss}}\_{\rho'}\to Y'.$$


**Corollary 6.** For every $k$-morphism, $\phi:X\to X',$ if the open subscheme $\phi^{-1}(X')^{\text{ss}}\_{\rho'}$ of $X$ contains $X^{\text{ss}}\_\rho$, and if the closed subscheme $\phi^{-1}(R\_{\rho'})\cap (X^{\text{ss}}\_\rho\times\_{\text{Spec}\ k}X^{\text{ss}}\_\rho)$ contains $R\_\rho$, then there exists a unique $k$-morphism $\phi\_Y:Y\to Y'$ such that $q'\circ \phi$ equals $\phi\_Y\circ q$.


**Proof.** By the first hypothesis, $q'\circ\phi$ is defined as a $k$-morphism $X^{\text{ss}}\_\rho\to Y'$. By the second hypothesis, $q'\circ \phi$ is invariant for the action $\rho$. Thus, since $q$ is a categorical quotient, there exists a unique $k$-morphism $\phi\_Y$ such that $\phi\_Y\circ q$ equals $q'\circ\phi$. **QED**


As a special case, note that the open subscheme $X^{\text{ss}}\_\rho\subset X$ and the closed subscheme $R\_q\subset X^{\text{ss}}\_\rho\times\_{\text{Spec}\ k}X^{\text{ss}}\_\rho$ depend only on the image of $\rho$, and they are also invariant under composing $\rho$ with $\gamma\_n$. Thus, the open subscheme $X^{\text{ss}}\_\rho$ and the $k$-morphism $q:X^{\text{ss}}\_\rho\to Y$ depend only on the image of $\rho$, and they are invariant under composing $\rho$ with $\gamma\_n$. (I am *not* saying that there are no other *interesting* structures coming from the group action. I am *only* saying that the quotient morphism depends only on the semistable locus and the relation.)


A more substantial example is as follows. Let $G$ be a reductive group $k$-scheme, and let $\rho$ be a morphism of group $k$-schemes, $$\rho:G\to \text{Aut}\_k(X,\mathcal{L}).$$ For each closed subgroup $k$-scheme $T$ of $G$ that is reductive, denote by $\rho\_T$ the restriction morphism of group $k$-schemes, $$\rho\_T:T\to \text{Aut}\_k(X,\mathcal{L}).$$ Let $T$ and $T'$ be closed subgroup $k$-schemes of $G$ that are both reductive. Denote the uniform categorical quotient $k$-morphisms by, $$q:X^{\text{ss}}\_{\rho\_T} \to Y\_T, \ \ q':X^{\text{ss}}\_{\rho\_{T'}} \to Y\_{T'}.$$ Let $g$ be a $k$-point of $G$ such that the conjugation, $$c\_g:G\to G, \ \ h\mapsto ghg^{-1},$$ restricts to an isomorphism from $T$ to $T'$. Denote by $r\_g$ the associated right translation, $$r\_g:X \to X, \ \ x \mapsto x\cdot \rho(g),$$ with its associated isomorphism $r\_g^\*\mathcal{L}\cong \mathcal{L}$.


**Corollary 7.** The $k$-isomorphism $r\_g$ mapso $X^{\text{ss}}\_{\rho\_T}$ isomorphically to $X^{\text{ss}}\_{\rho\_{T'}}$. The inverse image of the relation $R\_{T'}$ equals the relation $R\_T$. There is a unique $k$-morphism $r\_{g,Y}:Y\_T\to Y\_{T'}$ such that $r\_{g,Y}\circ q$ equals $q'\circ r\_g$.


**Proof.** This is a special case of Corollary 6. **QED**


**Question 8.** Let $k$ be an algebraically closed field. Let $T$ be a maximal torus in $\textbf{SL}\_n$. For the natural right action of $\textbf{SL}\_n$ on the Grassmannian $X\_{r,n}=\textbf{Grass}(r,n)$, with its unique linearization of the ample invertible sheaf $\omega\_{X\_{r,n}}^\vee$, for the associated geometric quotient $Y\_{r,n,T}=X\_{r,n}^{\text{ss}}/\rho(T)$, is there a $k$-isomorphism of $Y\_{r,n,T}$ with $Y\_{n-r,n,T}$?


For each $k$-point of $\text{Grass}(r,n)$, there is an associated $\textbf{SL}\_n$-equivariant isomorphism of $\text{Grass}(r,n)$ with the $k$-scheme $P\_r$ parameterizing parabolic subgroup schemes $H$ of $\textbf{SL}\_n$ in the same conjugacy class as the stabilizer of the $k$-point. The induced action of $\textbf{SL}\_n$ on $P\_r$ is by conjugation. 


There exists an **outer** automorphism (very much not unique), $$\phi:\textbf{SL}\_n \to \textbf{SL}\_n,$$ that sends every parabolic parameterized by $P\_r$ to a parabolic parameterized by $P\_{n-r}$. Thus, there is an induced $k$-isomorphism, $$\phi\_r: P\_r \to P\_{n-r}, \ \ H \mapsto \phi(H).$$ This $k$-isomorphism is certainly not $\textbf{SL}\_n$-equivariant, rather the two actions are intertwined by $\phi$, $$\phi( gHg^{-1}) = \phi(g)\phi(H)\phi(g)^{-1}.$$ For each maximal torus $T$ in $\textbf{SL}\_n$, denote by $T'$ the image maximal torus $\phi(T').$ Denote the uniform categorical quotients as follows, $$q:(P\_r)^{\text{ss}}\_T \to Y\_T, \ \ q':(P\_{n-r})^{\text{ss}}\_{T'} \to Y'\_{T'}.$$


**Corollary 9.** The $k$-isomorphism $\phi\_r$ restricts to a $k$-isomorphism, $$(P\_r)^{\text{ss}}\_T \to (P\_{n-r})^{\text{ss}}\_{T'}.$$ Moreover, the product $k$-isomorphism, $$(\phi\_r,\phi\_r):(P\_r)^{\text{ss}}\_T\times\_{\text{Spec}\ k} (P\_r)^{\text{ss}}\_T \to (P\_{n-r})^{\text{ss}}\_{T'}\times\_{\text{Spec}\ k} (P\_{n-r})^{\text{ss}}\_{T'},$$ maps the closed subscheme $R\_T$ isomorphically to the closed subscheme $R\_{T'}$. There is a unique $k$-morphism $\phi\_{r,Y}:Y\_T\to Y'\_{T'}$ such that $\phi\_{r,Y}\circ q$ equals $q'\circ \phi\_r.$


**Proof.** The follows from Corollary 6 in the same way as Corollary 7.
 **QED**


**Proposition 10.** There exists a $k$-isomorphism of $Y\_{r,n,T}$ with $Y\_{n-r,n,T}.$


**Proof.** By Corollary 7, there exists a $k$-isomorphism of $Y\_{n-r,n,T}$ with $Y\_{n-r,n,T'}.$ By Corollary 9, there exists a $k$-isomorphism of $Y\_{r,n,T}$ with $Y\_{n-r,n,T'}.$ **QED**


