I'm reviewing for exams and came across this problem from an older exam:
Let $G$ and $H$ be groups and let $\phi,\psi : H\to \operatorname{Aut}(G)$ be homomorphisms from $H$ into the group of automorphisms of $G$. If there exists $f\in \operatorname{Aut}(H)$ and $a\in\operatorname{Aut}(G)$ such that $\psi(f(h))=a\phi(h)a^{-1}$ as automorphisms of $G$ for all $h\in H$, then $G\rtimes_\phi H \cong G\rtimes_\psi H$.
I've proved this (unless I've made a mistake) by explicitly defining a map $F:G\rtimes_\phi H \to G\rtimes_\psi H$ by $(g,h)\mapsto(a(g),f(h))$ and checking that it is an isomorphism. However, choosing the map mostly involved guess-and-checking. That left me feeling like I somehow didn't do it "properly" and that I might have missed something I was supposed to notice or understand. I realize this is somewhat subjective, but I'm a bit suspicious, because the way I went about it makes the problem just seem like a strange, tedious way to test that someone knows the definition of a semidirect product.
Is there some other way to go about this problem (maybe using some clever argument about conjugation by $a$ inside the automorphism group?) that somehow illuminates or suggests something about the structure going on, or should it otherwise be obvious in some way what the isomorphism is? I guess my question is: is there something deeper going on here?