If $H$ and $K$ are non-isomorphic but of same order, can there be $\varphi_1, \varphi_2$ such that $N\rtimes_{\varphi_1}H\cong N\rtimes_{\varphi_2}K$? Let $N$, $H$ and $K$ be finite groups, such that $|H| = |K|$ and $H$ and $K$ being non-isomorphic. Can there exist $\varphi_1$ and $\varphi_2$ such that $ N \rtimes_{\varphi_1} H \cong N \rtimes_{\varphi_2} K $?
Naming $ G_1 := N \rtimes_{\varphi_1} H$ and $ G_2 := N \rtimes_{\varphi_1} K$, I know that $G_1$ and $G_2$ respectively contain copies of $H$ and $K$. I think in some special cases it is possible that say if $K \leqslant G_1$, then $G_1$ would have had a greater order because of containing both $H$ and $K$. Also I see that whatever $\varphi_1$ and $\varphi_2$ could be, they cannot both be trivial homomorphisms since then the semidirect product is a direct product and it's clear that $ N \times H \not\cong N \times K$.
But I wonder if in general we can prove that no such pair $\varphi_1$ and $\varphi_2$ exists for any arbitrary choice of $N$, $H$ and $K$.
 A: Let $N = \mathbf{Z}/3 \mathbf{Z}$, let $A = \mathbf{Z}/2 \mathbf{Z}$, and let $B = \mathbf{Z}/3
\mathbf{Z}$. If you take the non-trivial action of $A$ on $N$, you get
$$N \rtimes A = S_3.$$
Now take $H = A \times B$, where $A$ acts as above and $B$ acts trivially. Now clearly
$$N \rtimes H = S_3 \times \mathbf{Z}/3 \mathbf{Z}.$$
On the other hand, if $K = S_3$ acts trivially on $N$, then clearly
$$N \rtimes K = N \times K = \mathbf{Z}/3 \mathbf{Z} \times S_3 = N \rtimes H,$$
and $K$ is not isomorphic to $H$.
This is the simplest example of a more general construction. Let $A$ be a group, and let $V$ and $W$ be two representations of $A$ as vector spaces over $\mathbf{F}_p$ of the same dimension. That means that as abelian groups $V \simeq W \simeq N$ for some $N$. But now
$$V \rtimes (W \rtimes A) = (V \oplus W) \rtimes A = W \rtimes (V \rtimes A),$$
(here the action of $W \rtimes A$ on $V$ is the one factoring through the quotient to $A$ and similarly elsewhere) and now with $N=V=W$ and $H = W \rtimes A$ and $K = V \rtimes A$ one is done if $H$ and $K$ are not isomorphic. This will follow if $W$ and $V$ are "sufficiently different" representations of $A$. For example, if $p$ has order prime to $|A|$, then as representations they are not the same
under the action of the automorphism group of $A$ (which could include outer automorphisms). The example above corresponds to the group $A = \mathbf{Z}/2 \mathbf{Z}$ and its two different representations over $\mathbf{F}_3$.
