Let $H$ and $N$ be two groups with $H$ cyclic. Let $f,g:H \rightarrow \mathrm{Aut}(N)$ be homomorphisms such that $N\rtimes \_f H \cong N\rtimes \_g H$. Then does that mean $f(H)$ and $g(H)$ are conjugate in $\mathrm{Aut}(N)$?



Do isomorphic semi-direct products correspond to conjugate automorphisms?
The answer is no. You can easily have a situation where $f$ is the trivial map, while $g$ makes $H$ act through an inner automorphism of $N$, so that in both cases $N\rtimes H\cong N\times H$. For concreteness, let $N=D\_8$ be the dihedral group of order 8, let $\sigma$ be a non-central involution in $N$, let $H=\langle h\rangle\cong C\_2$, and define $g(h)(n)=\sigma^{-1} n\sigma\;\forall n\in N$. Then $G=N\rtimes\_g H\cong N\times H$, since the subgroup $\langle h\sigma\rangle$ of $G$ is of order 2, intersects $N$ trivially, and commutes with it.


