Let $G,H$ be groups. A homomorphism $\phi\colon G\longrightarrow \operatorname{Aut}(H)$ is an action such that:
$$\operatorname{Fix}(g)\le H, \forall g\in G \tag 1$$
where $\operatorname{Fix}(g):=\{h\in H\mid \phi_g(h)=h\}$$^{\dagger}$.
If $G$ and $H$ are both finite, say $G=\{g_1,\dots,g_{|G|}\}$ and $H=\{h_1,\dots,h_{|H|}\}$, then:
$$\sum_{i=1}^{|H|}|\operatorname{Stab}(h_i)|=\sum_{j=1}^{|G|}|\operatorname{Fix}(g_j)| \space\space\space\text{and (from Burnside's Lemma)}\space\space\space |G|\mid \sum_{j=1}^{|G|}|\operatorname{Fix}(g_j)| \tag 2$$
As an entry test for the possible utilization of $(2)$, let's take $G=C_p$ and $H=C_q$, where $p$ and $q$ are distinct primes. If a nontrivial homomorphism exists, then $|\operatorname{Fix}(g_{\bar j})|=1$ and $|\operatorname{Stab}(h_{\bar i})|=1$, for some $\bar j\in \{1,\dots,p\}$, $\bar i\in\{1,\dots,q\}$. Then $(2)$ yields:
$$[\space k+(q-k)p=l+(p-l)q\Longrightarrow k(p-1)=l(q-1)\space]\wedge [\space p\mid pq-l(q-1)\space ] \tag 3$$
for some $1\le k\le q$ and $1\le l\le p$. Now:
- if $p>q$, then from $(3)$-2nd term of the "$\wedge$": $p\mid pq-l(q-1)\Longrightarrow$ $p\mid l \Longrightarrow$ $l=p\Longrightarrow$ $k(p-1)=p(q-1)\Longrightarrow$ $k=\frac{p}{p-1}(q-1)>q-1\Longrightarrow$ $k=q$; but $(k,l)=(q,p)$ is not a solution of $(3)$-1st term of the "$\wedge$": so, for $p>q$, there are no nontrivial homomorphisms $\phi\colon C_p\longrightarrow\operatorname{Aut}(C_q)$;
- if $p<q$ and $p\nmid q-1$, then from $(3)$-2nd term of the "$\wedge$": $p\mid pq-l(q-1)\Longrightarrow$ $p\mid l$, and we fall back into the previous case.
Therefore, if $p\nmid q-1$, there are no nontrivial homomorphisms $\phi\colon C_p\longrightarrow\operatorname{Aut}(C_q)$. This hasn't used any knowledge about the isomorphism class of the group $\operatorname{Aut}(C_q)$. (Incidentally, that for $p\mid q-1$ there is actually a nontrivial homomorphism $\phi$, it is shown e.g. here.)
Can $(2)$/$(3)$ "framework" be used to exhibit other pairs $(G,H)$ such that only the trivial $\phi$ exists?
Edit. Let's now assume $G=C_p$ and $|H|=q^2$, still with $p,q$ distinct primes. If a nontrivial homomorphism exists, then $(2)$ yields:
$$[\space k+(q^2-k)p=l_0+l_1q+(p-l_0-l_1)q^2\space]\wedge [\space p\mid k\space ] \tag 4$$
for some $1\le k\le q^2$ and $1\le l_0+l_1\le p$. Since $p\mid k\Longrightarrow p\le k\le q^2$, for $p>q^2$ there are no nontrivial homomorphisms $\phi$. This suffices to give a different proof to, e.g., this question (and the conclusion doesn't change with $\Bbb Z_2\times \Bbb Z_2$ in place of $\Bbb Z_4$).
$^{\dagger}$In fact, for every $g\in G$, $\phi_g(1_H)=1_H$, whence $1_H\in\operatorname{Fix}(g)\ne\emptyset$; moreover, by definition, $\operatorname{Fix}(g)\subseteq H$; finally, for every $g\in G$ and $h_1,h_2\in\operatorname{Fix}(g)$, $\phi_g(h_1h_2^{-1})=\phi_g(h_1)\phi_g(h_2^{-1})=\phi_g(h_1)\phi_g(h_2)^{-1}=h_1h_2^{-1}$, whence $h_1h_2^{-1}\in\operatorname{Fix}(g)$.