Show that for $g \in G$ and $x \in X$ one has $\mathrm{Stab}_G(g*x)=g\,\mathrm{Stab}_G(x)g^{-1}$
Approach: Show
$$\mathrm{Stab}_G(g*x) \subseteq g\,\mathrm{Stab}_G(x)g^{-1}$$ $$g\,\mathrm{Stab}_G(x)g^{-1} \subseteq \mathrm{Stab}_G(g*x)$$
Let $p \in \mathrm{Stab}_G(g*x)$, so
$$p*(g*x)=g*x$$
$$(pg)*x=g*x$$
$$g^{-1}*(pg*x)=g^{-1}*(g*x)$$
$$g^{-1}pg*(x)=x,$$
so
$$g^{-1}pg \in \mathrm{Stab}_G(x)$$
which implies $$p \in g\,\mathrm{Stab}_G(x)g^{-1}$$
How does this look?
Let $p \in g\,\mathrm{Stab}_G(x)g^{-1}$, I am not quite sure how to handle this side. Can I just go backwards?
I find the other implication easier since $p \in gG_xg^{-1} \implies p*(g*x) = (ghg^{-1}g)*x = g*(h*x) = g*x$ where $h \in G_x$