I do not know much about Geometric Invariant Theory. My question is the following:
Let $X$ and $Y$ be two complex affine or projective varieties. Let $G$ be a reductive group which acts on both $X$ and $Y$. Let $\pi_X: X \to X//G$ and $\pi_Y: Y \to Y//G$ be the "projection" maps to the G.I.T quotients. If $Z\subset X$ and $W \subset Y$ are closed $G$-invariant subsets and $f: Z \to W$ is a bijection, under which conditions $f$ descends to a bijection $\bar{f}: \pi_X(Z) \to \pi_Y(W)$ ?
This kind of question is more easily adressed on the “other” side of algebraic geometry: let us look at what happens at level of function algebras.
Assume $f_0: X_0 \to Y_0$ is the corestriction of your morphism on a affine open subset $Y_0$ of $Y$ and a restriction to a $G$-stable affine open subset $X_0$ of $X$ mapped in $Y_0$ by $f$.
On the side of algebras, you have a morphism of algebras $A(f_0): A(Y_0) \to A(X_0)$ whose image is contained in the subalgebra $A(X_0)^G$ of $G$-invariant functions. So you can decompose $A(f_0) = A(\pi_0) \circ j$, where $j$ is induced by the canonical inclusion of $A(X_0)^G$ in $A(X_0)$ and define $\bar f$ by $A(\bar f) = j$.
In some situations, it might be impossible to find that $X_0$. Consider the case of the normaliser $N$ of a maximal torus $T$ in $\mathbf{SL}_2$ operating on $\mathbf{P}^1$: the element generating $N/T$ exchanges the two fixed points of $T$ so none of them is contained in a suitable $X_0$. This is why the theory of stable points used by Mumford has to remove these points before doing the quotient.
as long as $f$ passes to the quotient (i.e. sends orbits on orbits) it has the required property. Moreover, as Misha remarked, the semistable locus of the first quotient should be sent to the semistable locus of the second quotient. This basically mean that the pullback of the invariants on the codomain must give the invariants of the domain.
