Let $G$ be an algebraic group acting transitively on closed projective varieties $X$ and $Y$. Let $X \xrightarrow{f} Y$ be a bijective continuous map, that is $G-$equivariant. Does it follow that $X \to Y$ is an isomorphism of algebraic varieties?
It seems roughly intuitive because if the derivative of this map is zero somewhere then it should be zero everywhere by the equivariance.
If there is a transversality proof that is easier for $G$-manifolds and diffeomorphisms, or if it easier to prove it for complex algebraic varieties and biholomorphisms, i'd be all for it.
My motivation: There are two structures as a projective variety, that one can put on $GL(n)/B$ where $B$ is the Borel subgroup(which I define as a maximal connnected solvable subgroup of $GL(n)$).
Structure one: One can use Chevalley's theorem to define what $G/B$ means as a quasiprojective variety: its an orbit in $P(W)$ for some $G$ module $W$, which is therefore open in its closure. Then one can show that its closed $P(W)$ because $B$ is solvable, and orbits of minimal dimension are closed. Therefore $GL/B$ is a closed orbit in $P(W)$.
Structure two: There is a bijection of $G-$sets between $GL(n)/B$ and $Flag(\mathbb{C}^n)$. There is a closed embedding of this into a product of grassmanians, and grassmanians are closed subvarieties of projective space by plucker. So $GL(n)/B$ can be regarded as a closed subspace of $P(\wedge^1 \mathbb{C^n} \otimes ...\wedge^{n-1} \mathbb{C^n})$.
I want to know 'Why are these $G$-varieties isomorphic as projective varieties?' and the question above would allow me to answer this.
What I have got from professors: A professor at tea told me that I need to use the correspondence between maps from projective space and line bundles over the domain. Brainstorming: Maybe I need to use something about $G$ equivariant line bundles over $G/B$ since these will correspond to $G$-equivariant maps from $G/B$.. Borel-weil-bott theorem anyone?
I am not closely familiar with algebraic groups, but the corresponding question for Lie groups has a simple postive answer: Take a point $x_0\in X$, and denote by $G_{x_0}$ its stabilizer in $G$. Then $g\mapsto g\cdot x_0$ induces a diffeomorphism $G/G_{x_0}\to X$. Since $f$ is equivariant and bijective, you see that for $y_0=f(x_0)$, you get $G_{y_0}=G_{x_0}$ and $g\mapsto g\cdot y_0$ induces a diffeomorphism $G/G_{y_0}\to Y$, so $X$ and $Y$ are $G$-equivariantly diffeomorphic.
On the other hand, for your motivating example, the two constructions for $G/B$ you describe are essentially the same. The natural choice for the representation $W$ in Chevalley's construction is that the highest weight of $W$ is the sum of all fundamental weights. The natural construction of $W$ is a the $G$-invariant subspace in the tensor product of all fundamental representations. These fundamental representations are just $\mathbb C^n$, $\Lambda^2\mathbb C^n$, ..., $\Lambda^{n-1}\mathbb C^n$. So $P(W)$ is naturally a subspace of $P(\mathbb C^n\otimes\dots\otimes \Lambda^{n-1}\mathbb C^n)$. By definition, the embedding of $G/B$ into $P(W)$ is induced by $g\mapsto g\cdot w_0$, where $w_0$ is a highest weight vector in $W$. The natural choice for $w_0$ is the tensor product of the highest weight vectors in the fundamental representations. But you can also obtain the Plücker embedding of the Grassmannian $Gr(m,\mathbb C^n)$ similarly to Chevalley's construction, but starting from the highest weight vector in $\Lambda^m\mathbb C^m$. So you readily see that the embedding you construct via PlÜcker actually has values in $P(W)$ and coincides with the Chevalley embedding.