Find all group of order $20$ which is a semidirect product of a cyclic group of order $4$ by a cyclic group of order $5$ Problem: Find all group of order $20$ that are a semidirect product of a cyclic group of order $4$ by a cyclic group of order $5$.

My attempt: We knew that a cyclic group of order $n$ is isomorphic to $\mathbb{Z}_n$, so we have $C_4 \cong \mathbb{Z}_4$, $C_5 \cong \mathbb{Z}_5$ and $\mid C_4 \mid = \mid \mathbb{Z}_4 \mid = 4$, $\mid C_5 \mid = \mid \mathbb{Z}_5 \mid = 5$. Then we find a semidirect product of $\mathbb{Z}_4$ by $\mathbb{Z}_5$. To find this, we define an automorphism $\theta \colon \mathbb{Z}_5 \rightarrow \text{Aut}(\mathbb{Z}_4)$, while $\text{Aut}(\mathbb{Z}_n) = \{\sigma_k \mid k \in \mathbb{Z}, (k,n)=1\}$, so $\text{Aut}(\mathbb{Z}_4) = \{\sigma_1,\sigma_3\}$ and $\mid \text{Aut}(\mathbb{Z}_4) \mid = 2$.

My professor told that there is only the trivial homomorphism $\mathbb{Z}_5 \rightarrow \text{Aut}(\mathbb{Z}_4)$ since $(5,2)=1$.

So my question: 


  Why from $(5,2)=1$ can we conclude that there is only one trivial homomorphism, and for the trivial homomorphism, how do I define it?
Why from $(5,2)$ we can conclude that there is only one trivial homomorphism


Let $f:\Bbb Z_5 \to \text{Aut}(\Bbb Z_4) \sim \Bbb Z_2$ be a homomorphism with $f(1)=a$. Now order of  $f(1)$ divides both $5$ and $2$. But their gcd is one, so $a$ must be identity and hence $f$ is trivial!
$$ 
\DeclareMathOperator{\Ker}{Ker} 
\DeclareMathOperator{\Im}{Im}
\DeclareMathOperator{\Aut}{Aut}
$$
Also, there's another way to prove $f$ is trivial. Consider corollary from homomorphism theorem:
$$f: G → H, \quad\textrm{where $f$ is a homomorphism}$$
$$|G| = |\Im f|\cdot|\Ker f| $$ So in our case $5 = |\Im f| |\Ker f|$. We know that $|\Im f| &lt;= |\Aut(Z_4)| = |\Aut(Z_2)| = 2$. Thus, the only possible way is that $|\Ker f| = 5 $ and $|\Im f| = 1$, which concludes $f$ is trivial homomorphism.