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 $p$-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.

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.