Actions of G on corresponding orbits are equivalent for stable maps

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

$\Rightarrow$

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 $Y$,

$h: X \rightarrow Y$

and

$h(Gx)=f(gx)\, \forall g\in G=g.f(x)\, \forall g\in G = h(Gy)$

thus each orbit of $X$ corresponds one to one with an orbit of $Y$.

$\Leftarrow$

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.

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I assume stable means the same thing as $G$-equivariant. (I also assume your question is for feedback on your proofs; you don't explicitly state any question.) In your $\Rightarrow$ direction you seem a bit trigger-happy with your '$\forall g\in G$'s. I can't follow what you're saying in the $\Leftarrow$ direction. When in doubt, use plain English over symbols. – anon Feb 6 '13 at 6:29
Stable means $f(g.x)=g.f(x)$ for all g. I am asking for help on the proof, yes. The $\Leftarrow$ is underdeveloped. I had a few ideas going at once and wasn't careful with the editing. I'll try to make it more concise. – zzzzzzzzzzz Feb 6 '13 at 6:36