A question about normal subgroup $H<G,aHa^{-1}<G$. Then, $H$ is isomorphic to $aHa^{-1}$. I want to show $aha^{-1}\in H,\forall h\in H,a\in G$, but I cannot figure it out. Any hint is appreciated.

To show that $H\cong gHg^{-1}$, you need to show that for any $a\in G$, the conjugation $$\kappa_a : G\to G,\quad g\mapsto aga^{-1}$$ is an automorphism of $G$. The proof is easy: For $g,h\in G$ we have $$\kappa_a(gh) = agha^{-1} = ag(a^{-1}a)ha^{-1} = (aga^{-1})(aha^{-1}) = \kappa_a(g)\kappa_a(h)\text{,}$$ so $\kappa_a$ is a homomorpisms. To show that $\kappa_a$ is one-to-one, check $\kappa_a\circ\kappa_{a^{-1}} = \operatorname{id}$.

Now every subgroup $H$ is isomorphic to $\kappa_a(H) = aHa^{-1}$.