Let $G$ and $H$ be groups, and $\varphi\colon G\to\operatorname{Aut}(H)$ a homomorphism. Then, further than all the results valid for a general group action on a set, the following additional one holds: $$\operatorname{Fix}(g):=\{h\in H\mid \varphi_g(h)=h\}\le H \tag 1$$ For finite $G$ and $H$, the condition $(1)$ brings new opportunities, as now $\left|\operatorname{Fix}(g)\right|$ divides $|H|$; as an example, for $p, q$ distinct primes such that $p\nmid q-1$, one can prove from here that there are no nontrivial homomorphisms $\phi\colon C_p\to\operatorname{Aut}(C_q)$, by no means knowing anything on the structure of $\operatorname{Aut}(C_q)$. If, in addition, $\varphi$ is injective, then this setting may lead to some other result: for instance, for $q$ prime and $H=C_q$, I was able to prove via $(1)$ that $|G|$ divides $q-1$, again by no means knowing anything on the structure of $\operatorname{Aut}(C_q)$$^\dagger$.
Question. One step ahead in this escalation, if $\varphi$ is an isomorphism, I expect this setting to lead to further group action conditions (involving stabilizers, fixed points subgroups, orbits, etc.), potentially useful in getting the structure (isomorphism class) of $\operatorname{Aut}(H)$, for some known $H$. In particular, as entry test of this approach, I'm aiming to retrieve in this way the known result $\operatorname{Aut}(C_q)\cong C_{q-1}$.
Edit (2021-11-5). The integer $|G|-k$ is the number of $g\in G$ such that $\left|\operatorname{Fix}(g)\right|=q$, namely the number of $g\in G$ such that $\varphi_g(h)=h$ for every $h\in C_q$, namely the number of $g\in G$ such that $\varphi_g=\operatorname{Id}_{C_q}$, whence: $$|G|-k=\left|\operatorname{ker}\varphi\right| \tag{1bis}$$ If $\varphi$ is injective, then $\left|\operatorname{ker}\varphi\right|=1$ and $(1\text{bis})$ yields: $k=|G|-1$. But $k$ is the number of $g\in G$ such that $\left|\operatorname{Fix}(g)\right|=1$, namely the number of $g\in G$ such that $\varphi_g(h)=h\Longrightarrow h=1$. Therefore, every nontrivial $g\in G$ is sent to an automorphism of $C_q$ which moves all the nontrivial elements of $C_q$. If, in addition, $G\cong\operatorname{Aut}(C_q)$, then every $\psi\in\operatorname{Aut}(C_q)\setminus\{\operatorname{Id}_{C_q}\}$ moves all the nontrivial elements of $C_q$. So, $C_q$ has at most $q-1$ automorphisms, all but one (the identity) of which move all the nontrivial elements of $C_q$. Suppose we have proved that there are precisely $q-1$ automorphisms: would the fact that $q-2$ of them move all the nontrivial elements of $C_q$ imply that some of them (automorphisms) must have order $q-1$?
Edit (2021-11-8). By the previous edit, every $\psi\in\operatorname{Aut}(C_q)\setminus\{\operatorname{Id}_{C_q}\}$ is of the form: \begin{alignat}{1} &\psi(1)=1 \\ &\psi(a^i)=a^{\sigma(i)} \\ \end{alignat} where $\sigma\in S_{q-1}$ has cycle type $(r_1,\dots,r_N)$ ($r_1\le\dots\le r_N$), for some $N\ge 1$, with:
- $r_i\ge 2$, for every $i=1,\dots,N$
- $\sum_{i=1}^Nr_i=q-1$
Note that, if a cycle $(i_1\dots i_s)$ which composes $\sigma$ is such that $\sum_{j=1}^si_j\not\equiv 0\pmod q$, then: \begin{alignat}{1} \psi(a^{i_1}\dots a^{i_s}) &= \psi(a^{i_1+\dots+i_s\pmod q}) \\ &= a^{\sigma(i_1+\dots+i_s\pmod q)} \\ \end{alignat} and: \begin{alignat}{1} \psi(a^{i_1})\dots\psi(a^{i_s}) &= a^{\sigma(i_1)}\dots a^{\sigma(i_s)} \\ &= a^{\sigma(i_1)+\dots+\sigma(i_s)\pmod q} \\ \end{alignat} whence: \begin{alignat}{1} a^{\sigma(i_1+\dots+i_s\pmod q)} &= a^{\sigma(i_1)+\dots+\sigma(i_s)\pmod q} \\ \end{alignat} and finally (being the $a^i$'s distinct and by definition of cycle): \begin{alignat}{1} \sigma(i_1+\dots+i_s\pmod q) &= \sigma(i_1)+\dots+\sigma(i_s)\pmod q \\ &= i_1+\dots+i_s\pmod q \\ \end{alignat} which is a contradiction, because $\sigma$ doesn't fix any element of $\{1,\dots,q-1\}$. Therefore, every cycle $(i_1\dots i_s)$ which composes $\sigma$ must fulfil the condition: $$\sum_{j=1}^si_j\equiv 0\pmod q\tag{1ter}$$ Maybe , the condition $(1\text{ter})$, which is fulfilled in particular by the $(q-1)$-cycles, limits the possibilities enough that, for some of the nontrivial automorphisms $\psi_1,\dots,\psi_{q-2}$, the corresponding $\sigma$ must be a $(q-1)$-cycle?
Edit (2021-11-29). By taking as known that$^{\dagger\dagger}$ $\operatorname{Aut}(C_q)\cong (\Bbb Z/q\Bbb Z)^\times$, we can take advantage of $G$ being a finite abelian group, and use the framework in the previous Edit to prove, e.g., that $\operatorname{Aut}(C_7)$ (and then $(\Bbb Z/7\Bbb Z)^\times$) is cyclic. In fact, the order of every element divides the maximal order among the elements in $G$. By contradiction, let's assume that such maximal order is $3$. Then, all the nontrivial elements must have order $3$. But then, the only nontrivial permutations fulfilling the constraints in the previous Edit are: $\sigma_1=(124)(356)$, $\sigma_2=(142)(356)$, $\sigma_3=(124)(365)$, $\sigma_4=(142)(365)$, namely too few to build up the whole $\operatorname{Aut}(C_7)$. Likewise, let's assume that such maximal order is $2$. Then, all the nontrivial elements must have order $2$. But then, the only nontrivial permutation fulfilling the constraints in the previous Edit is $\sigma_1=(16)(25)(34)$, definitely too few to build up the whole $\operatorname{Aut}(C_7)$. Therefore, $\operatorname{Aut}(C_7)$ (or, equivalently, $(\Bbb Z/7\Bbb Z)^\times$) must have an element of (maximal) order $6$. Maybe this kind of argument can be generalized to every $q$?
$^\dagger$Every automorphism of $C_q$ is a permutation of its $q-1$ nontrivial elements; therefore, $\operatorname{Aut}(C_q)\cong K\le S_{q-1}$ and hence $\left|\operatorname{Aut}(C_q)\right|$ divides $(q-1)!$. For an embedding $\varphi\colon G\hookrightarrow\operatorname{Aut}(C_q)$, $|G|$ divides $\left|\operatorname{Aut}(C_q)\right|$ and $\operatorname{Fix}(g):=\{h\in C_q\mid \varphi_g(h)=h\}$ is a subgroup of $C_q$. But $C_q$ has no nontrivial subgroups; therefore: \begin{alignat}{1} \sum_{g\in G}\left|\operatorname{Fix}(g)\right| &= k+(|G|-k)q \\ &= |G|q-k(q-1) \\ \tag 2 \end{alignat} for some $k$, $0\le k\le |G|$. By Burnside's Lemma, $|G|$ divides the LHS of $(2)$, and hence $|G|$ divides $k(q-1)$ either. The case $k=|G|$ corresponds to a transitive action, whence $q\mid |G|$: contradiction, because $|G| \mid\left|\operatorname{Aut}(C_q)\right|$ and $\left|\operatorname{Aut}(C_q)\right|\mid(q-1)!$, but $q\nmid (q-1)!$. Therefore, $k<|G|$ and then necessarily $|G|\mid q-1$.
$^{\dagger\dagger}$After all, here the focus is to prove the cyclicity of $\operatorname{Aut}(C_q)$, and the standard result $\operatorname{Aut}(C_q)\cong U_q$ can be then assumed as known - see e.g. Herstein's Topics in Algebra, 2nd Edition, Example 2.8.1, page 69.