Consider a group $G$ acting on a set $X$. Suppose $g,h,a∈G$ such that $h=aga^{−1}$. Prove that the formula $x↦ax$ defines a bijection of sets $Fix_X(g)→Fix_X(h)$.
I know that $Fix_X(g)=\{x \in X: gx=x\}$. And correct me if I'm wrong, but it is obviously surjective because for any $y \in Fix_X(h)$ by just taking $a^{-1}y$ to get the value in $Fix_X(g)$. However, I don't know how prove that it is injective. So is this the correct approach (proving surjectivity and injectivity) and how would I prove injectivity?
Yes, a map is a bijection if and only if it is surjective and injective, and yes your method of showing surjectivity of the map $x\mapsto ax$ is correct. Furthermore, injectivity of the map $x\mapsto ax$ isn't too difficult: If $x,y\in Fix_X(g)$ and $ax=ay$, then $$x=a^{-1}(ax)=a^{-1}(ay)=y.$$