If $N$ is a normal subgroup of $G$ and $H$ a characteristic subgroup of $N$ (i.e., $\sigma(H)=H$ for all $\sigma \in \operatorname{Aut}(N)$). Show that $H$ is normal in $G$.
Does one needs to show that $\sigma(gHg^{-1})=gHg^{-1}$ for every g in G. Would that imply H is normal? is there a alternative way?
On
The other answerer seems to have missed the subtlety that $H$ is not necessarily characteristic in $G$, only in $N$. However, an inner automorphism of $G$ leaves $N$ invariant and induces an automorphism on $N$. Since $H$ is characteristic in $N$, $H$ is also left invariant, hence $H$ is normal in $G$.
Normal subgroups are by definition exactly those subgroups which are invariant under inner automorphisms, that is automorphisms of the form $x \mapsto g x g^{-1}$ for some fixed $g \in G$. Thus a characteristic subgroup, i.e. a subgroup which is invariant under all automorphims, is trivially normal.