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.

Any automorphism of $\mathbb{Z}_p$ preserves whether an element is divisible by $p^k$, so it is Lipschitz (in particular, continuous) with respect to the $p$-adic norm. On the other hand, any automorphism must preserve $\mathbb{Z}$, which is dense in $\mathbb{Z}_p$.

What I should've said is that any automorphism is determined by its behavior on $\mathbb{Z}$, hence by its behavior on $1$. To make up for that mistake, let me offer a sketch of the description of the structure of $\mathbb{Z}_p^{\ast}$. This group clearly splits up as the direct product of $(\mathbb{Z}/p\mathbb{Z})^{*}$ and the multiplicative group $U = 1 + p \mathbb{Z}_p$. It is now an interesting exercise to show that the exponential map $x \mapsto u^x, x \in \mathbb{Z}, u \in U$ extends to a map from $\mathbb{Z}_p$ to $U$ which, given the right choice of $u$, is an isomorphism for $p > 2$. For $p = 2$, Yiftach Barnea's otherwise excellent answer is slightly wrong and $U$ is in fact isomorphic to $\{ \pm 1 \} \times \mathbb{Z}_2$.