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Let $X$ be a topological space and $G$ a group acting on $X$. Do we have this property:

$$\operatorname{orb}(x)=\operatorname{orb}(y)\iff\operatorname{stab}(x)\sim \operatorname{stab}(y)\qquad ?$$ where $\operatorname{orb}(x)$ is the orbit of $x$, $\operatorname{stab}(x)=\{g\in G\mid gx=x\}$, and the symbol $\sim$ means conjugate to.

One way is obvious: if $\operatorname{orb}(x)=\operatorname{orb}(y)$, then $x=gy$ for some $g\in G$, so $$\operatorname{stab}(x)=\operatorname{stab}(gy)=g\operatorname{stab}(y)g^{-1}.$$ But the other way is not obvious to me: if $\operatorname{stab}(x)\sim \operatorname{stab}(y)$, then $\exists k\in G$ such that for all $g\in \operatorname{stab}(x)$, $ \exists h \in \operatorname{stab}(y)$ such that $g=khk^{-1}$. Now since $gx=x$ and $hy=y$, then $khk^{-1}x=x$ so $hk^{-1}x=k^{-1}x$ hence $h\in \operatorname{stab}(k^{-1}x)$, but I can't go any further... Thanks for your help.

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    Is there some hypothesis you've left out? This is certainly falsifiable (as written) by considering a trivial group action in which every group element fixes every point.2011-12-04
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    At least $G$ needs to act faithfully.2011-12-04
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    This approach seems quite optimistic. Forgetting about the topology for a second, if you examine any action of say the integers on an uncountable set, there will be uncountably many orbits but only countably many conjugacy classes of stabilizers. More generally, you are trying to capture global dynamical behavior by a local, algebraic criterion, which is a massive loss of information.2011-12-04

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