Isomorphism of two non-abelian groups of order $pq$ Let $p$ and $q$ be two primes such that $q\mid p-1$. Suppose $\phi, \varphi$ are two non-trivial homomorphism from $\mathbb{Z}_q$ to $Aut(\mathbb{Z}_p)$. How to define an isomorphism from $\mathbb{Z}_p\rtimes_\phi \mathbb{Z}_q$ to $\mathbb{Z}_p\rtimes_\varphi \mathbb{Z}_q$. Thanks for any help.
 A: Take $\phi_1$ and $\phi_2$ to be two morphism from $\mathbb{Z}_q$ to $Aut(\mathbb{Z}_p)$, then it is clear that $\phi_1$ and $\phi_2$ are uniquely defined by the image of $1$ (a generator of $(\mathbb{Z}_q,+)$). Then because $Aut(\mathbb{Z}_p)=(\mathbb{Z}_p^*,\times)$ $\phi_1(1)$ and $\phi_2(1)$ are just given by the image of $1$ in $\mathbb{Z}_p^*$. Here you need to understand that the following is an isomorphism :
$$Aut(\mathbb{Z}_p)\rightarrow  (\mathbb{Z}_p^*,\times)$$
$$\psi\mapsto \psi(1)$$
The inverse morphism being $\lambda\mapsto [ a\mapsto \lambda.a]$. 
Now because $\phi_1(1)$ and $\phi_2(1)$ cannot be trivial by hypothesis (both are non-nul elements of $\mathbb{Z}_p$) hence they are both generators of the same cyclic subgroup in $Aut(\mathbb{Z}_p)$ ($Aut(\mathbb{Z}_p)$ is cyclic). Now take $\psi:\mathbb{Z}_q\rightarrow \mathbb{Z}_q$ such that $\phi_1(1)=\phi_2(\psi(1))$ then you have $\phi_1(b)=\phi_2(\psi(b))$. 
$$\mathbb{Z}_p\rtimes_{\phi_1}\mathbb{Z}_q\rightarrow  \mathbb{Z}_p\rtimes_{\phi_2}\mathbb{Z}_q$$
$$(a,b)\mapsto (a,\psi(b))$$
is a group isomorphism. Let us see that :
$$f((a,b)(c,d))=f(a+\phi_1(b).c,b+d)=(a+\phi_1(b).c,\psi(b)+\psi(d))$$
$$f(a,b)f(c,d)=(a,\psi(b))(c,\psi(d))=(a+\phi_2(\psi(b)).c,\psi(b)+\psi(d)) $$
Hence we have the equality by $\phi_1(b)=\phi_2(\psi(b))$.
