Let us recall this fact. Let $G$ be a semisimple algebraic group over $\mathbb C$ and let $V,V'$ be two irreducible $G$-representations. We denote by $X,X'$ the unique closed $G$-orbits contained in $\mathbb P V, \mathbb P V'$ respectively. We know that if
$$
\mathbb P V \supset X \cong X' \subset \mathbb P V'
$$
as projective $G$-varieties, then $\mathbb PV \cong \mathbb PV'$ as projective spaces. In particular, $\dim V=\dim V'$.


I want to understand the inverse direction: if I have two irreducible $G$-representations $W,W'$ of the same dimension, should I conclude that the closed $G$-orbits $Y \subset \mathbb P W, Y' \subset \mathbb P W'$ are isomorphic as projective $G$-varieties?



Do representations of same dimension implies isomorphic closed orbits?
No, this is not true. For instance the symplectic group
$$
G = \mathrm{Sp}(V),
$$
where $V$ is a symplectic vector space of dimension 6
has two irreducible representations of dimension 14:
$$
V\_1 = \wedge^2V^\vee / \langle \omega \rangle,
\qquad\text{and}\qquad 
V\_2 = \wedge^3V^\vee / (V^\vee \wedge \omega),
$$
where $\omega$ is the symplectic form. These representations are not isomorphic, as well as the corresponding closed orbits, which are the isotropic Grassmannians
$$
X\_1 = \mathrm{IGr}(2,V)
\qquad\text{and}\qquad 
X\_2 = \mathrm{IGr}(3,V).
$$


By the way, the inverse is also not true unless the line bundles associated with the embeddings of $X \to \mathbb{P}(V)$ and $X' \to \mathbb{P}(V')$ are identified by the isomorphism $X \cong X'$.


