If I write two papers one will basically be just a combinatorial argument and an algebraic topology argument. I think I'd have to wait to post the first one until I did the second because the arguments would be too easy to scoop.
Assume K is finite cyclic group, H is an arbitrarily subgroup. Suppose $\phi_1$ and $\phi_2$ are homomorphism from $K \rightarrow Aut(H)$ such that $\phi_1(K)$ and $\phi_2(K)$ are conjugate subgroups in Aut(H). I want to prove that $H \rtimes_{\phi_1} K$ and $H \rtimes_{\phi_2} K$ are isomorphic subgroups.
The way I did this is as follows
So we know K is cyclic right so it is generated by some element k i.e K = <k>. By Hypothesis $\sigma \phi_1(k) \sigma^{-1} = \phi_2(k)^a$ for some $a \in \mathbb{Z}$
so we can define the following homomorphism
$(h,k) \mapsto (\sigma(h),k^a)$
I proved that this is a group homomorphism between $H \rtimes_{\phi_1} K$ to the other one.
Since K is finite so it has some order n. Since $\phi_1(K)$ and $\phi_2(K)$ are conjugate so they have same order m, they are also cyclic.
oh ok I got it now @BalarkaSen since we know that this particular a is relatively prime mod n so we know it is invertible i.e we must have b such that ab = 1.