I've been trying to solve the following question, but am unable to fully complete the required proof.
Let $p$ be a prime number, $C$ a cyclic group of order $p$ and $N$ a finite group such that $Aut(N)$ has a normal $p$-Sylow subgroup of order $p$. Prove that for every non-trivial $\phi_1, \phi_2 : C \to Aut(N)$, we have: $N \rtimes_{\phi_1} C \cong N \rtimes_{\phi_2} C$.
We may assume $C$ is $\mathbb{Z}_p$ since: $N \rtimes_{\phi_1} C \cong N \rtimes_{\phi_1} \mathbb{Z}_p \cong N \rtimes_{\phi_2} \mathbb{Z}_p \cong N \rtimes_{\phi_2} C $.
Since a normal $p$-Sylow subgroup is unique, we know that the only options for homomorphisms $\phi_i$ are defined by:
$$ \phi_i : \mathbb{Z}_p \to Aut(N)$$ $$ 1 \mapsto \alpha^i$$
where $\alpha$ is an automorphism of order $p$ from the unique $p$-Sylow subgroup. There are no other homomorphisms, since, if there are, they must be of order $p$ (being non-trivial) and therefore there is another $p$-Sylow subgroup of order $p$ -- contradiction.
Now I want to prove that for every such $\phi_1, \phi_2$ we have: $$ N \rtimes_{\phi_1} \mathbb{Z}_p \cong N \rtimes_{\phi_2} \mathbb{Z}_p $$ but because of the presence of the semi-direct product, I'm not sure how to construct such an isomorphism. How does one approach such a construction?
Without loss of generality we may assume that $\phi_1(1) = \alpha$ and so there exists some $n \in \{1,\dots,p-1\}$ such that $\phi_2(1) =\alpha^n$. Since $C_p$ is cyclic, these values entirely determine $\phi_1,\phi_2$. One way to think of this is the the $n$-th power map is an isomorphism between the image of $\phi_1$ and $\phi_2$. Then we want to use this map to construct an isomorphism between the semidirect products. Let $m$ be the inverse of $n$ modulo $p$, i.e. $mn \equiv 1 \mod p$. Consider the map $\psi:N \rtimes_{\phi_1} C_p\to N \rtimes_{\phi_2} C_p$ defined by $$\psi(g,i) = (g,mi).$$ Hopefully it is clear that this is a bijection so it just remains to show that it is a homomorphism. For this we can see that$$\psi((g,i)(h,j)) = \psi(g\alpha^i(h),i+j) = (g\alpha^i(h),m(i+j)).$$ On the other hand, $$ \psi(g,i)\psi(h,j) = (g,mi)(h,mj) = (g{(\alpha^{n})}^{mi}(h),mi+mj) = (g\alpha^{mni}(h),m(i+j)) = (g\alpha^i(h),m(i+j)).$$ Note that in this second computation we are working with the product in $N \rtimes_{\phi_2} C_p$, hence we have a factor of $n$ appearing in the power of $\alpha$.