Are orbits equal if and only if stabilizers are conjugate?

Question

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.

Answer 1

The answer is no.

Let the trivial group $G=\{Id\}$ act on $X$ containing at least two different points $x$ and $y$. Then $\operatorname{orb}(x) \neq \operatorname{orb}(y)$ but $\operatorname{Stab}(x)=\operatorname{Stab}(y)$.

More generally, invariant points (i.e. with stabilizer equal to $G$) are not in the same orbit but have the same stabilizer.