I am trying to prove that given some group $G$ which acts on elements in $A$ where elements $g\in G$ and elements $a,b\in A$ (where $b=g.a$) that the stabilizer of $b$ is equal to the conjugate of the stabilizer of $a$.
$ \text{Stab}(b) = g.\text{Stab}(a).g^{-1} $
I know that the stabilizer can be considered as sets and for the sets to be equal they have to be subsets of each other. So I have got to the point of trying to prove both sides of this and expression, $\text{Stab}(b) \subseteq g.\text{Stab}(a).g^{-1} \wedge g.\text{Stab}(a).g^{-1}\subseteq \text{Stab}(b)$.
And I know for the sets to be subsets to be equal that all elements in the left have to be a member of the right.
So considering the first case of the and ($\wedge$). I have $$ \vdash\forall s\in \text{Stab}(b),\ s\in g.\text{Stab}(a).g^{-1} $$ And unfolding the definition of being in the stabilizer (not sure about this step) $$ \vdash s.b=b, \implies g.s.a.g^{-1} =a $$ Substituting in the value of $b$ $$ \vdash s.(g.a)=(g.a), \implies g.s.a.g^{-1} =a $$ And I am a bit stuck at this step, as the group isn't known to be abelian. Haven't got much further proving the case the other way.
I am overcomplicating it? Have I made an error in my current steps? Is there a simpler method?