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 $(mathbb{Z}/pmathbb{Z})^{*}$ 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 ${ pm 1 } times mathbb{Z}_2$.
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