Question- If $G$ is complete, then the holomorph of $G$ is isomorphic to $G$×$G$.
I am studying semidirect products for the first time, and in some notes i found this exercise. As far as i know about this problem, if $G$ is a group then let $H$=Aut($G$) and let $\phi$:$H \to Aut(G) $ be the identity map, i.e. all elements go to itself, then Holomorph of $G$ is $G \rtimes_\phi Aut(G)$. Now if $G$ is complete then its outer automorphism group is identity, then $G \cong Aut(G)$ so Holomorph is $G \rtimes_\phi G$ but $G \rtimes_\phi G \cong G\times G$ when $\phi$ is the trivial homomorphism i.e. everything goes to 1.
so what am i missing here?