TITLE: When are two semidirect products isomorphic?
QUESTION [35 upvotes]: Let $N$, $H$ be groups, and $\varphi : H \to \operatorname{Aut}(N)$ a group homomorphism. Then we can form $N \rtimes_{\varphi} H$, the semidirect product of $N$ and $H$ with respect to $\varphi$. 

Is there a way to determine if $\varphi_1, \varphi_2 : H \to \operatorname{Aut}(N)$ lead to isomorphic semidirect products by just comparing $\varphi_1$ and $\varphi_2$?

REPLY [7 votes]: There is a theorem by Taunt (in "Remarks on the Isomorphism Problem in the Theories of Construction of Finite Groups"), which gives a nice characterization if $\gcd(|N|,|H|)=1$. I'm citing from "Construction of Finite Groups" by Besche and Eick.

Let $N,H$ be finite (soluble) groups with $\gcd(|N|,|H|)=1$. Furthermore let $\psi_i:H\rightarrow Aut(N)$ for $i=1,2$ be two homomorphisms. Define $G_i:=N\rtimes_{\psi_i}H$.
Then $G_1\cong G_2$, iff there exists automorphisms $\alpha\in Aut(N)$ and $\beta\in Aut(K)$ such that $(h^\beta)^{\psi_2}=(h^{\psi_1})^\alpha$ for all $h\in H$.
(i.e. $(\psi_2(\beta(h))=\alpha^{-1}\psi_1(h)\alpha\quad \forall h\in H$ )

The solubility is not necessary, therefore you can omit it.