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 $|C_4| = |\mathbb{Z}_4| = 4$, $|C_5| = |\mathbb{Z}_5| = 5$. Then we find a semidirect product of $\mathbb{Z}_4$ by $\mathbb{Z}_5$. To find this, we define an automorphism $\theta \colon \mathbb{Z}_5 \rightarrow \mathrm{Aut}(\mathbb{Z}_4)$, while $\mathrm{Aut}(\mathbb{Z}_n) = \{\sigma_k \mid k \in \mathbb{Z}, (k,n) = 1\}$, so $\mathrm{Aut}(\mathbb{Z}_4) = \{\sigma_1, \sigma_3\}$ and $|\mathrm{Aut}(\mathbb{Z}_4)| = 2$.
My professor told that there is only the trivial homomorphism $\mathbb{Z}_5 \rightarrow \mathrm{Aut}(\mathbb{Z}_4)$ 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?

 A: 
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!
A: A way to prove that the homomorphism $f\colon \Bbb Z_5\to\operatorname{Aut}(\Bbb Z_4)$ is only the trivial one without any prior knowledge of the structure of $\operatorname{Aut}(\Bbb Z_4)$ (not even its order), is the following.
Since $\Bbb Z_5$ is of prime order, action's pointwise stabilizers are trivial or the whole $\Bbb Z_5$. Therefore:
$$\sum_{i=0}^3\left|\operatorname{Stab}(i)\right|=k+5(4-k)\tag1$$
for some integer $k$, $0\le k\le 4$. By Burnside's (counting) lemma, $5$ must divide the RHS of $(1)$, whence $k=0$: all the stabilizers do coincide with the whole $\Bbb Z_5$, and so does the kernel of the action. Therefore, the action is trivial.

This argument can be developed further to prove that, for $p$ and $q$ distinct primes such that $p\nmid q-1$, the only homomorphism $f\colon \Bbb Z_p\to\operatorname{Aut}(\Bbb Z_q)$ is the trivial one (so, again without any prior knowledge of the structure of $\operatorname{Aut}(\Bbb Z_q)$).
With reference to the action $f$, we get:
$$\sum_{i=0}^{q-1}\left|\operatorname{Stab}(i)\right|=\sum_{j=0}^{p-1}\left|\operatorname{Fix}(j)\right| \tag2$$
where in this case $\operatorname{Fix}(j)$ is a subgroup of $\Bbb Z_q$, for every $j$. By the primality of the orders of $\Bbb Z_p$ and $\Bbb Z_q$, $(2)$ and Lagrange yield:
$$k+p(q-k)=l+q(p-l) \tag3$$
for some integers $k,l$ such that $0\le k\le q-1$ and $0\le l\le p-1$. From $(3)$:
$$k(p-1)=l(q-1) \tag4$$
By Burnside's (counting) lemma, $p\mid k$ (see the LHS of $(3)$); but since by assumption $p\nmid q-1$, from $(4)$ follows $p\mid l$: contradiction, because $p>l$. So, if $p\nmid q-1$, there isn't any nontrivial action (by automorphisms) of $\Bbb Z_p$ on $\Bbb Z_q$.
A: $$ 
\DeclareMathOperator{\Ker}{Ker} 
\DeclareMathOperator{\Im}{Im}
\DeclareMathOperator{\Aut}{Aut}
$$
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 $|\Im f| <= |\Aut(Z_4)| = |Z_2| = 2$. Thus, the only possible way is that $|\Ker f| = 5 $ and $|\Im f| = 1$, which concludes $f$ is trivial homomorphism.
