If $H$ is a subgroup of $G$, then if $N(H) = \{g \in G~|~gHg^{-1}=H\} ; ~~C(H) = \{g \in G ~|~ ghg^{-1}=h ~\forall~h \in H\}$, then I have proved that :
$\Psi: N(H) \rightarrow \operatorname{Aut}(H)$ given by $\Psi(n) = nhn^{-1} ~\forall~h \in H,~ n \in N(H)$ is a homomorphism and its kernel is $C(H)$
Hence, by the first isomorphism theorem,
$N(H)/ C(H) ~\approx~ \Psi[N(H)]$
I am confused how is $\Psi[N(H)]$ a subgroup of $\operatorname{Aut}(H)$ as per the statement of the $N/C$ theorem.
I know that $Inn(H)$ is a subgroup of $\operatorname{Aut}(H)$
Is $\Psi[N(H)]$ a subgroup of $\operatorname{Inn}(H)$ by any chance?
Help will be appreciated. Thank you.