Given actions of G on X and on Y, these actions are equivalent if and only if there is a bijection from $X $ \ $ G \rightarrow Y$ \ $G$ so that actions of G on corresponding orbits are equivalent.
Def. An action of G on X and Y is equivalent if there is a stable bijection $f: X \rightarrow Y$.
My Attempt at a proof
Suppose $f:X\rightarrow Y$ is stable, then choose an $x\in X$ and let the corresponding orbit be $Gx$. Then we have a map from the set of orbits of X to the set of orbits of Y,
$h: X \rightarrow Y$
$h(Gx)=f(gx) \forall g\in G=g.f(x) \forall g\in G = h(Gy)$
thus each orbit of X corresponds one to one with an orbit of Y.
Well we know each orbit of X is mapped to a corresponding orbit of Y. So we take the map $h$ from orbit to orbit and then we show that for each x in that orbit (since orbits are disjoint or equal) we can move the g's in G to outside without affecting the y's.