Some context: I'm trying to find $\text{Aut}(\mathbb{Z}_n)$ for any natural number $n$ (Problem I.2.15c in Hungerford, if you're wondering), and I'm using the Chinese Remainder Theorem to deduce that if $n=p_1^{r_1}\dots p_k^{r_k},$ then $\mathbb{Z}_n \cong \mathbb{Z}_{p_1^{r_1}} \times \dots\times \mathbb{Z}_{p_k^{r_k}},$ and since $\text{Aut}(\mathbb{Z}_n)$ has the same order as $U(\mathbb{Z}_n)$ (I'm thinking they're isomorphic, too, but I can't think of a way to prove it), I'd think $U(\mathbb{Z}_n)\cong U(\mathbb{Z}_{p_1^{r_1}}) \times\dots\times U(\mathbb{Z}_{p_k^{r_k}}).$ That's the source of my question. From there, I know $U(\mathbb{Z}_{p_i^{r_i}})$ has order $p_i^{r_i} - p_i^{r_i-1}$, but I'm hard pressed to determine general structure. I know $U(\mathbb{Z}_{p_i^{r_i}})$ is a product of cyclic groups (since it's abelian) of orders of powers of $p$ with the last cyclic group of the product having order $p^j (p-1)$ , but I have no idea what else I can say about them specifically.
Thanks so much!