On isomorphic normal subgroups of a group

Question

Well this is quite stupid a question because the intuition is false. See comments and answers below.

Original question:

Consider $H_1,H_2\le G$, $H_1 \cong H_2$. It seems trivial that if $H_1 \unlhd G$, then $H_2 \unlhd G$, but how can I give a rigorous proof? I'm stuck after I write $\varphi(gH_1g^{-1})=\varphi(H_1)=H_2 $ because I don't know whether $\varphi$ can be an automorphism (so that all elements in $G$ can be written as $\varphi(g)$ for some $g$).

Answer 1

It seems trivial that if $H_1 \unlhd G$, then $H_2 \unlhd G$

You mean trivially false. ;)

Consider any group $G$ with a subgroup $H\subseteq G$ which is not normal. Then consider $G\times H$. With this we have that $\{e\}\times H$ is normal in $G\times H$, but $H\times\{e\}$ is not, even though they are isomorphic.