If $N$ is a normal subgroup of $G$, and we have an isomorphism $\phi \colon G \to H$, how can I show that $\phi(N)$ is a normal subgroup of $H$?
It seems intuitive, but the only thing I can think of for a proof is saying that the homomorphism property shows that $\phi(gng^{-1}) = \phi(g)\phi(n)\phi(g)^{-1} \in \phi(N)$
You're almost there. In particular, it suffices to show that for all $h \in H, n \in N,$ we have $h \phi(n) h^{-1} \in \phi(N)$. With that in mind, take your equation $$ \phi(gng^{-1}) = \phi(g)\phi(n)\phi(g)^{-1} \in \phi(N) $$ and set $g = \phi^{-1}(h)$, noting that $\phi$ is an isomorphism.