Let $G$ be an affine algebraic group (let's say over $\mathbb{C}$). If necessary one can assume $G$ to be reductive. Imagine one has $X$ over which $G$ acts freely: moreover, we have a locally closed subvariety $Y$ such that $X=G \cdot Y$.


Moreover, one has a subgroup $H \subseteq G$ such that $H$ stabilizes $Y$. We know also that given any two points $y\_1,y\_2 \in Y$ and $g \in G$ such that $g y\_1=y\_2$ we have $g \in H$.


As everything acts freely one can make up two quotient varieties $Y/H$ and $X/G$ with the natural morphism $$f:Y/H \to X/G .$$


The hypotheses imply that $f$ is bijective. Is it also an isomorphism? I'm not assuming anything on normality or smoothness



Quotient variety and subgroups
I don't think that this is true without extra hypotheses like normality of $X$ (if $X$ is normal then it follows as explained by Damian Rossler in the comments). Here is a possible example.


We will take $G = GL\_2$ and $H=1$ (the latter is just the trivial group). Let $C$ be the cuspidal curve $C = Spec(k[x,y]/(y^2-x^3))$.


Let's take $X = GL\_2 \times C$, equipped with the natural action of $GL\_2$ by right multiplication on the first factor (this is just the trivial $GL\_2$-bundle over $C$).


Inside $GL\_2$ we have the additive group $\mathbb{G}\_a$ of strictly upper triangular matrices. So we have a closed subvariety $Z = \mathbb{G}\_a \times C$ inside $X$. We can write $Z = Spec(k[x,y,t](y^2-x^3))$. Inside $Z$ we can define $Y$ to be the vanishing locus of the polynomial $xt-y$. So we get a closed subvariety $Y \hookrightarrow X$ given by


$$ Y = Spec(k[x,y,t]/(y^2-x^3, xt-y))$$


Notice that $X/GL\_2 =C$ and the projection $Y \to C$ onto the second component is the map $Y/H \to X/GL\_2$ in this case.


$Y$ is actually isomorphic to $\mathbb{A}^1$, and the projection $Y \to C$ exhibits $Y$ as the normalization of the cuspidal curve $C$. In particular $Y \to C$ is not an isomorphism.


Note that the projection $Y \to C$ is bijective on points, so $Y$ contains exactly one point in each fiber of the $GL\_2$-bundle $X \to C$. This implies that $GL\_2 \cdot Y = X$ and also that there are no two closed points of $Y$ that belong to the same $GL\_2$-orbit. In other words, the hypotheses are satisfied for $Y \hookrightarrow X$ (with $H = 1$) and the map $Y/H \to X/G$ is not an isomorphism.


One can probably do variations of this to get examples with $H$ nontrivial.


