Let $G$ be a group acting on a set $X$. If two elements in $X$ belong to the same orbit (under the action of $G$), then their stabilizers $H_x$, $H_y$ are conjugate in $G$.
I would just like to show this particular point: if $g \in H_x$ then $h.g.h^{-1} \in H_y$.
I wrote the following proof but I think it's wrong and I don't know where is my mistake.
Let $g \in H_x$. Then $x=g.x \Longleftrightarrow h.y=g.(h.y)=gh.y$.
Then: $y=e.y=h^{-1}h.y=h^{-1}.(h.y)=h^{-1}.(gh.y)=(h^{-1}gh).y$ so we conclude that $h^{-1}gh \in H_y$ and $g \in hH_yh^{-1}$.
Any help would be appreciated.
If $x,y$ belong to the same orbit, then $gx=y$ for some $g\in G$. We have $H_x = \{h\in G\mid hx=x\}$.
Now $h\in H_y$ iff $y\in H_{gx} = \{h\mid h(gx) = (hg)x=gx\} = \{h\mid (g^{-1}hg)x=x\}$ iff $g^{-1}hg\in H_x$.
It follows that $H_x=g^{-1}H_yg$.