Let $G = N \rtimes H$ be the semidirect product of two cyclic groups $N$ and $H$, and assume that there exists a cylic group $K \leq G$ such that $G \simeq N \rtimes K$. Is it true that $H$ and $K$ are conjugate in $G$ by an element of $N$?
I feel this should be true and that it should have a very straightfoward demonstration. However, all I've managed to prove so far is that both $H$ and $K$ act on $N$ through the same automorphism of $N$, that is, if $H = \langle h\rangle$ and $K = \langle k \rangle$ then $$hnh^{-1} = knk^{-1}\,,\quad \forall n \in N\,.$$
I was mostly interested in the infinite cyclic case $N \simeq H \simeq K \simeq \mathbb{Z}$, but I also feel that this should not really depend on the isomorphism classes of the subgroups involved.
After playing with this for a while, I think the following is a counter-example:
Let $$G = \mathbb{Z}/4\mathbb{Z} \rtimes_\varphi \mathbb{Z}/2\mathbb{Z} = \langle x, y \mid x^4 = y^2 = 1\,, yxy^{-1} = x^3\rangle\,,$$ and take $N = \langle x \rangle$, $H = \langle y \rangle$ and $K = \langle xy \rangle$. We have both $G \simeq N \rtimes H$ and $G \simeq N \rtimes K$. However, any conjugate of $y$ by $x^i \in N$ satisfies: $$x^iyx^{-i} = x^ix^{-3i}y = x^{-2i}y\,.$$
Hence, $K$ is not a conjugate of $H$ by an element of $N$.