Let the group $G$ act on the set $X$. Suppose $x,y\in X$ and that we have $y=g*x$. Prove that $h\in stab_G(x)$ if and only if $ghg^{-1}\in stab_G(y)$.
Attempt: By definition $stab_G(x)=\{g\in G:g*x=x\}$. So $ghg^{-1}\in stab_G(y)$ means that $(ghg^{-1})*y=y$. Since this is a group action, we have $(gh)*g^{-1}y=y$, that is, $(gh)*g^{-1}gx=y$. Therefore $(gh)*x=y$. I don't know how this verifies that $h\in stab_G(x)$ though, can anyone give any tips?
The notation $stab_G(x)$ is cumbersome here since there is no possible ambiguity on the group $G$. So I will simply write $stab(x)$. You're almost there. $y=g*x$
$$\begin{align} (ghg^{-1})*y=y&\iff (ghg^{-1})*(g*x)=g*x\\ &\iff (gh)*x=g*x\\ &\iff h*x=e*x=x \end{align}$$
by multiplication by $g^{-1}$ of the two members of the preceding equality($G$ is a group, any element $g$ admits an inverse)
So, $$(ghg^{-1})*y=y \iff h\in stab(x).$$
Q.E.D.