Is there a pattern in classifying $\operatorname{Aut}(\mathbb{Z}_n)$ as $\mathbb{Z}_k$ or direct sums of such groups? If $n$ is prime, then $\operatorname{Aut}(\mathbb{Z}_n)\sim \mathbb{Z}_{n-1}$. If $n$ is not prime however, then a bunch of different things seem to happen. I computed them manually for $n\leq 14$. We have
- $\operatorname{Aut}(\mathbb{Z}_4)\sim \mathbb{Z}_2$
- $\operatorname{Aut}(\mathbb{Z}_6)\sim \mathbb{Z}_2$
- $\operatorname{Aut}(\mathbb{Z}_8)\sim \mathbb{Z}_2\bigoplus\mathbb{Z}_2$
- $\operatorname{Aut}(\mathbb{Z}_9)\sim \mathbb{Z}_6$
- $\operatorname{Aut}(\mathbb{Z}_{10})\sim \mathbb{Z}_4$
- $\operatorname{Aut}(\mathbb{Z}_{12})\sim \mathbb{Z}_2\bigoplus\mathbb{Z}_2$
- $\operatorname{Aut}(\mathbb{Z}_{14})\sim \mathbb{Z}_6$
Is there a pattern here that I'm missing? To me this looks like complete randomness.
You need a few tools to classify $\operatorname{Aut}(\mathbb{Z}_n)$:
(1) If $H$ and $K$ are groups of relatively prime order, then $\operatorname{Aut}(H\times K)\cong\operatorname{Aut}(H)\times \operatorname{Aut}(K)$.
(2) $\operatorname{Aut}(\mathbb{Z}_n)\cong U(n)$ where $U(n)$ is the group of units modulo $n$.
By the Chinese Remainder Theorem, if $n=p_1^{\alpha_1}\dots p_k^{\alpha_k}$ where $p_i$'s are primes, then $$\operatorname{Aut}(\mathbb{Z}_n)\cong U(p_1^{\alpha_1})\times\dots\times U(p_k^{\alpha_k}).$$
The following results can be found in An Introduction to The Theory of Groups by Rotman, Theorem 5.44 and Theorem 6.7:
$$U(p^n)\cong\begin{cases}\mathbb{Z}_{p^{n-1}(p-1)} &\text{ if } p \text{ is odd}\\ \mathbb{Z}_2\times \mathbb{Z}_{2^{n-2}} &\text{ if } p=2 \text{ and }n\geq 3\end{cases}$$