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.