Prove $\operatorname{stab}_G(g \cdot x) = g\operatorname{stab}_G(x)g^{-1}$ Suppose a group $G$ acts on a set $X$. The stabilizer in $G$ of $x \in X$ is $$ \operatorname{stab}_G(x) = \\{a \in G : a \cdot x = x \\} $$ For each $g \in G$, let $$ g\operatorname{stab}_G(x)g^{-1} = \\{gag^{-1} : a \in \operatorname{stab}_G(x)\\} $$ I have to prove that $$ \operatorname{stab}_G(g \cdot x) = g\operatorname{stab}_G(x)g^{-1} $$ by showing that * $\operatorname{stab}_G(g \cdot x) \subseteq g\operatorname{stab}_G(x)g^{-1}$ * $g\operatorname{stab}_G(x)g^{-1} \subseteq \operatorname{stab}_G(g \cdot x) $

Let $a\in\operatorname{stab}_G(g\cdot x)$; then $a\cdot(g\cdot x)=g\cdot x$, that is $$ (ag)\cdot x=g\cdot x $$ or else $$ (g^{-1}ag)\cdot x=x $$ Therefore $b=g^{-1}ag\in\operatorname{stab}_G(x)$, which means that $$ a=gbg^{-1} $$ where $b\in\operatorname{stab}_G(x)$. This proves one of the inclusions; can you show the other one with a similar technique?