Let $H$ and $N$ be two groups, $\phi$ and $\psi$ be two homomorphisms $H \to \operatorname{Aut}(N)$, and $G_1 = N \rtimes_\phi H$ and $G_2 = N \rtimes_\psi H$ corresponding semidirect products.
What are conditions on $\phi$ and $\psi$ for $G_1 \cong G_2$?
If for some automorphisms $h \in \operatorname{Aut}(H)$ and $n \in \operatorname{Inn}(\operatorname{Aut}(N))$ we have
$$\phi = n \circ \psi \circ h\tag1$$ then $G_1 \cong G_2$.
At least for infinite groups, this condition is not necessary: let $X^\omega$ be direct sum of countable number of copies of $X$, let $N = S_3^\omega \times \mathbb Z_3^\omega \times \mathbb Z_3^\omega$ and $H = \mathbb Z_2^\omega \times \mathbb Z_2^\omega$, then trivial homomorphism and homomorphism that sends
$i$-th copy of $\mathbb Z_2$ of the first component of $H$ to non-trivial automorphism of $i$-th copy of $\mathbb Z_3$ of the second component of $N$, and sends all elements from the second component of $H$ to trivial automorphism,
both give group $\mathbb S_3^\omega \times \mathbb Z_6^\omega$, but there are no automorphisms on $\operatorname{Aut}(N)$ and $H$ that transforms one of this homomorphisms to the other.
Is condition $(1)$ necessary for finite groups? Is there some "good" general condition?
Best Answer
Your criterion for equivalence of semidirect products corresponds to group isomorphisms that preserve the distinguished subgroup $N$ of the semidirect product. You would expect to find examples in which the two semidirect products were isomorphic under an isomorphism that mapped $N$ to a different normal subgroup.
I did a computer calculation (in Magma) and found that the group $\mathtt{SmallGroup}(128,753)$ of order $128$ has, up to equivalence under its automorphism group, two types of normal subgroup $N$ isomorphic to $C_2^4$ (i.e. elementary abelian of order $16$) with complements isomorphic to $D_8$ (dihedral of order $8$).
One of them has $2$ conjugacy classes of complements and the other has $8$, and they are not equivalent as semidirect products according to your criterion.