Sorry if this is an easy one, I'm a little rusty on my group theory. My first guess was that it's simply the inverse limit of the Aut($\mathbb{Z}/p^i\mathbb{Z})$, with the map when $i\leq j$ given by taking $\sigma\in$ Aut$(\mathbb{Z}/p^j\mathbb{Z})$ to the map $\tilde{\sigma}:\mathbb{Z}/p^i\mathbb{Z}\rightarrow\mathbb{Z}/p^i\mathbb{Z}$ defined by solving $\phi\circ\sigma=\tilde{\sigma}\circ\phi$ where $\phi:\mathbb{Z}/p^j\mathbb{Z}\rightarrow\mathbb{Z}/p^i\mathbb{Z}$ is the reduction map, but that seems too optimistic - I couldn't think of any reason $\tilde{\sigma}$ would be well-defined, much less be an automorphism of $\mathbb{Z}/p^i\mathbb{Z}$.


Also, barring a full answer to my question, I would be interested in whether Aut$(\mathbb{Z}\_p)$ is a $p$-group. If not, what can we say about the elements $\sigma\in$ Aut$(\mathbb{Z}\_p)$ with order a power of $p$?



What is the automorphism group of the additive group of the p-adic integers?
First the $p$-adic integers are finitely generated (actually cyclic) pro-$p$ group therefore from a result of Serre all automorphisms are continuous. Now as it cyclic it is enough to see what happens to $1$. It has to go to another generator, i.e. any element of the form $a\_0+a\_1p+a\_2p^2+\cdots$, where $0 \leq a\_i < p$ for all $i$ and $a\_0 \ne 0$. Hence, every element in $\mathbb{Z}\_p$ is just multiplied by $a\_0+a\_1p+a\_2p^2+\cdots$. Thus, the automorphism group is the multiplicative group of $\mathbb{Z}\_p$.


It is of course not a pro-$p$ group, but it contains a subgroup of index $p-1$ which is pro-$p$ and is actually again a cyclic pro-$p$ group, i.e. isomorphic to $\mathbb{Z}\_p$ and thus have no elements of finite order.


