Equivariant homomorphism in a non-abelian group of order $pq$, where $p$ and $q$ are distinct primes. Let $G$ be a group and $H$ be a normal subgroup of $G$. A homomorphism $\phi$ from $H$ to $G$ is said to be $G$-equivariant if $\phi (ghg^{-1})=g \phi (h)g^{-1}$, for all $g \in G$ and $h \in H$. I wants to compute all $G$-equivariant homomorphisms from $\mathbb{Z}_p$ to a non-abelian group $G$ of order $pq$, where $p$ and $q$ are distinct primes with $p>q$. I have tried this by property of homomorphisms, if we restrict ourself to $\mathbb{Z}_p$ in the codomain then all homomorphisms are $G$-equivariant. But after that, I am not able to proceed. Please help me.

Consider $G = \langle a\rangle:\langle b\rangle\cong \mathbb{Z}_p:\mathbb{Z}_q$, a semidirect product, by $bab^{-1} = a^{m}$ for some $m$ such that $m^q\equiv 1\pmod p$. Let $\phi:\langle a\rangle\to G$ be a homomorphism, then $\phi(a) = a^n$ for some $n$, because $\phi(a)$ has order $1$ or $p$.

For any $g\in G$, $g = a^kb^l$ for some $k,l$. Then $gag^{-1} = a^kb^lab^{-l}a^{-k} = a^{m^l}$, and so $\phi(gag^{-1}) = a^{m^ln}$, while $g\phi(a)g^{-1} = a^kb^la^nb^{-l}a^{-k} = a^{m^ln}$. So they are equal.