$\DeclareMathOperator\SL{SL}\DeclareMathOperator\Aut{Aut}$EDITED Let $G=\SL\_{n,{\mathbb{C}}}$, the special linear group over ${\mathbb{C}}$.
Let $H\subset G$ be a finite subgroup.
Set $X=G/H$ be the corresponding homogeneous space, it is a complex variety.
Let $\tau\in \Aut({\mathbb{C}})$ be an automorphism of the field of complex numbers,
not necessarily continuous.
Consider the conjugate variety $\tau X$.


We have $\tau X=\tau G/\tau H$.
We take the standard $\mathbb{Q}$-form $\SL\_{n,\mathbb{Q}}$ of $G$, then we identify $\tau G$ with $G$,
and we obtain $\tau X= G/\tau(H)$.


**Question 1.** Is it always true that $\tau X\simeq X$ as homogeneous spaces of $G=\SL\_{n,\mathbb{C}}$?


**Question 2.** Is it always true that $\tau X\simeq X$ as algebraic varieties over ${\mathbb{C}}$?


**Question 3.** Is it always true that that the $(\tau X)({\mathbb{C}})\simeq X({\mathbb{C}})$ as $C^\infty$-manifolds?


I expect the answer NO to Question 1.  

Let us try to construct a counter-example.
We wish to construct a finite subgroup $H\subset G$ such that $G/\tau(H)\not\simeq G/H$,
i.e., $\tau(H)$ is not conjugate to $H$ in $\SL(n,\mathbb{C})$.
We come to the following question:


**Question 4.** Let $\rho\colon H\to \SL(n,\mathbb{C})$ be an $n$-dimensional complex representation of a finite group $H$.
Is it possible that $\tau(\rho(H))$ is not conjugate to $\rho(H)$ in $\SL(n,\mathbb{C})$ for some $\tau\in \Aut({\mathbb{C}})$?


In the last question by $\tau(\rho(H))$ we mean $\{ \tau(\rho(h))\, |\ h\in H\}$,
where $\tau(\rho(h))$ is the matrix with matrix elements $\tau(\rho(h)\_{i,j})$.



Conjugation of the quotient of $\mathrm{SL}(n,\mathbb{C})$ by a finite subgroup
$\DeclareMathOperator\SL{SL}
\DeclareMathOperator\GL{GL}
\DeclareMathOperator\PSL{PSL}
\DeclareMathOperator\Aut{Aut}
\DeclareMathOperator\Out{Out}
\DeclareMathOperator\Ad{Ad}
$I can only address Question 1/Question 4. Let me replace $\SL\_n$ with $\GL\_n$; our $H$ will be simple, so it will provide an example for $\SL\_n$ also.
If $\rho, \psi: H \to \GL\_n(\mathbb C)$ are faithful representations with the same image, then $\psi^{-1} \circ \rho$ is an automorphism of $H$. Hence asking whether $(\tau\rho)(H)$ is conjugate to $\rho(H)$ as a set for faithful $\rho$ is asking whether there is an automorphism $\varphi \in \Aut(H)$ such that $\tau \rho$ is isomorphic to $\rho \varphi$ as a complex representation of $H$. This reduces the problem to character theory. Note also that the action of $\Aut(H)$ on characters factors through $\Out(H)$.


An example is given by $H = \PSL\_2(\mathbb F\_p)$ for $p > 13$.
For any nonsquare $a \in \mathbb F\_p^\times$,
$$\Out(H) = \left\langle \Ad \begin{pmatrix} a & 0 \\ 0 & 1\end{pmatrix}\right\rangle \cong \mathbb Z/2\mathbb Z.$$
See [1] or [2] for this.
Let
$$T = \left\{ \begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix} \mid x \in \mathbb F\_p^\times \right\} / \{\pm 1\}$$
be the diagonal torus in $H$.
If $\zeta$ is a $(p-1)/2$th root of unity, let $\alpha: \mathbb F\_p^\times \to \mathbb C^\times$ be a multiplicative character sending a primitive element of $\mathbb F\_p^\times$ to $\zeta$. Note that $\alpha(-1) = 1$.
There is an irreducible representation of degree $p+1$ given by parabolically inducing the character $\begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix} \mapsto \alpha(x)$ of $T$ up to $H$. Its character $\chi$ satisfies
$$ \chi\begin{pmatrix} x & 0 \\ 0 & x^{-1}\end{pmatrix} = \alpha(x) + \alpha(x^{-1}).$$
See e.g. [3] for a description of the characters of $\mathrm{PSL}\_2(\mathbb F\_p)$. If $\varphi$ is the automorphism of conjugation by $\begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}$, then $\varphi$ fixes $T$ elementwise, so $\chi\varphi$ has the same values on $T$ as $\chi$.



> 
> Lemma: If $\zeta$ is a primitive $m$th root of unity, then $\zeta^{-1} + \zeta$ is irrational for $m \notin \{1,2,3,4,6\}$.
> 
> 
> 


This lemma also appears in the theory of root systems.


If $\zeta$ is chosen to be a primitive $(p-1)/2$th root of unity, then $\zeta + \zeta^{-1}$ is irrational for $(p-1)/2 \notin \{1,2,3,4,6\}$, i.e. $p = 11$ or $p > 13$. This provides an example of a character $\chi$ and $\tau \in \Aut(\mathbb C)$ such that $\tau \chi \neq \chi \varphi$ for all $\varphi \in \Out(H)$.




---


Proof of the lemma: We may assume $m > 2$.
The degree $[\mathbb Q(\zeta):\mathbb Q]$ is $\phi(m)$.
As $\zeta + \zeta^{-1}$ is real, $[\mathbb Q(\zeta): \mathbb Q(\zeta + \zeta^{-1})] > 1$, but $\zeta$ satisfies the quadratic equation
$$\zeta^2+1=(\zeta+\zeta^{-1})\zeta $$
over $\mathbb Q(\zeta + \zeta^{-1})$, so $[\mathbb Q(\zeta):\mathbb Q(\zeta + \zeta^{-1})] = 2$. Hence the degree of $\zeta^{-1} +\zeta$ is $\phi(m)/2$. If $m = p\_1^{k\_1}\cdots p\_\ell^{k\_\ell}$, then $\phi(m) = p\_1^{k\_1 - 1}(p\_1 - 1)\cdots(p\_\ell^{k\_\ell})(p\_\ell - 1)$. If $\phi(m) = 2$, then primes $p\_i$ appearing must have $p\_i - 1$ divides $2$, whence $p\_i$ can be only $2$ or $3$. Further, only $m=3$, $m=4$, and $m=6$ are possible.




---


**References**


[1] Steinberg, R. (1960). Automorphisms of Finite Linear Groups. Canadian Journal of Mathematics, 12, 606-615. doi:10.4153/CJM-1960-054-6


[2] [What is the outer automorphism group of $\operatorname{SL}(2,\mathbb{F}\_q)$?](https://mathoverflow.net/questions/348440/what-is-the-outer-automorphism-group-of-operatornamesl2-mathbbf-q/348441#348441)


[3] Bonnafé, *Representations of $SL\_2(\mathbb F\_q)$*, §5.3, or Fulton and Harris, *Representation Theory*, §5.2


