This is a followup to this previous question: Do homomorphisms $H \to \operatorname{Aut}(K)$ that coincide at the level of $\operatorname{Out}(K)$ induce isomorphic semidirect products?
I am trying to understand why $S_n$ doesn't seem to appear as a normal subgroup of a nontrivial semidirect product except when $n=6$. Two relevant facts: $S_n$ is centerless when $n \geq 3$, and its outer automorphism group is trivial except when $n = 6$. This raises the more general question:
If $\phi,\psi:H \to \operatorname{Aut}(K)$ induce the same map $H \to \operatorname{Out}(K)$, and if $K$ is centerless, then is it true that $K \rtimes_\phi H \cong K \rtimes_\psi H$?
Answer
The answer is still no.
As an example, let $K=A_5$ and $H = \langle a,b\mid a^4=b^2=1,ab=ba \rangle = C_4 \times C_2$.
Choose $\phi$ such that $\phi(a)$ is conjugation by $(1,2,3,4)$ and $\phi(b)=1$.
Choose $\psi$ such that $\psi(a)$ is conjugation by $(1,2)$ and $\psi(b)$ is conjugation by $(1,2)(3,4)$.
Then $\phi$ and $\psi$ induce the same map to ${\rm Out}(K)$, but the associated semidirect products are not isomorphic, essentially because $\phi$ and $\psi$ have different kernels. In fact the one with $\phi$ has centre a Klein $4$-group and the one with $\psi$ has cyclic centre, which is similar to the example I gave for the previous question.
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