I feel like I did something wrong, because I never actually used $a \in G$.
Proof: Define $\alpha: H \rightarrow a^{-1}Ha$ by $\alpha(h)=a^{-1}ha$
Let $x,y \in H$ then,
$\alpha(xy) = a^{-1}xya\\ =(a^{-1}xa)(a^{-1}ya)\\ =\alpha(x) \alpha(y)\\$
So this implies that $\alpha$ is a homomorphism.
Injective: Let $x,y \in H$ such that $\alpha(x)= \alpha(y)$
Then $a^{-1}xa=a^{-1}ya$, multiply on left and right by $a$ and $a^{-1}$
$aa^{-1}xaa^{-1}=aa^-1yaa^{-1}$
$x=y$ This implies $\alpha$ is one-to-one.
Surjective: Let $x \in a^{-1}Ha$. Then $x=a^{-1}ha$ for some $h \in H$.
$\alpha(h)=a^{-1}ha=x$
So $\alpha$ is onto.
Therefore, $H $ and $a^{-1}Ha$$ is an isomorphism.