Assume that K is a cylic group, H is an arbitrary group and $\phi_{1}$ and $\phi_2$ are homomorphisms from K into Aut(H) such that $\phi_1(K)$ and $\phi_2(K)$ are conjugate subgroups of Aut(H). If K is infinite assume $\phi_1$ and $\phi_2$ are injective. Prove by constructing an explicit isomorphism that $H \rtimes_{\phi_1} K \cong H \rtimes_{\phi_2} K$.
I am having problems in the infinite case and for the finite case I have no idea how to do it.
Here is what I did suppose $\sigma$ does this conjugation i.e $\sigma \phi_1(K) \sigma^{-1} = \phi_2(K)$, so we get since K is cyclic we get $\sigma\phi_1(k)\sigma^{-1} = \phi_2(k)^a \ \forall\ k \in K$. I showed that the map defined as $\psi : H \rtimes_{\phi_1} K \rightarrow H \rtimes_{\phi_2} K$, such that $(h,k) \mapsto (\sigma(h),k^a)$ is a homomorphism now I need to construct the 2-sided inverse. I have no idea how to do it in finite or infinite case.
Well, only one such $a$ can describe it all!
Take the generator element $k$ of the cyclic group. Then $\sigma\phi_1(k)\sigma^{-1}=\phi_2(k)^a$, as you said, for some $a\in\Bbb Z$. But then, for any element $k^n$ of $K$, we have $$\sigma\phi_1(k^n)\sigma^{-1}=\phi_2(k)^{an}\,.$$
Now, for the infinite case, $\,a$ must be $\pm1$, else we wouldn't get the whole $\phi_2(K)$ by conjugating $\phi_1(K)$ by $\sigma$.
For the finite case, the condition implies $|\ker\phi_1|=|\ker\phi_2|$, and as $K$ is cyclic, we must have $\ker\phi_1=\ker\phi_2$, and thus we can reduce the problem to the injective homomorphisms $\phi_i': K/\ker\phi_i\to {\rm Aut}(H)$.
Say, the order of this cyclic quotient group is $m$.
Finally, the above argument shows that $a$ must be relatively prime to $m$.