The $p$-adic numbers $\Bbb Q_p$ also have trivial automorphism group. The proof starts out in the same way as for $\Bbb R$, and then you have to show $\varphi(z)=z$ for $z\in\Bbb Q$ implies the same for $z\in\Bbb Q_p$. For this, you have to show that any automorphism $\varphi$ of the field is automatically continuous. I suppose there are many methods of doing this, but one way is to look at the set $S$ defined in the following purely algebraic way: an element $s\in\Bbb Q_p$ is in $S$ if and only if for every $m$ prime to $p$, $1+s$ has an $m$-th root in $\Bbb Q_p$.
Notice that such $s$ must necessarily be in $\Bbb Z_p$, the integers of the field. Furthermore, such $s$ can not be a unit of $\Bbb Z_p$, because either $1+s\in p\Bbb Z_p$, and has only finitely many roots in $\Bbb Q_p$, or $1+s$ is also a unit (necessarily then $p>2$), and doesn’t have a $(p-1)^2$-th root in $\Bbb Q_p$, because its class modulo $p$ doesn’t even have such a root.
On the other hand if $s\in p\Bbb Z_p$, then $1+s$ is a principal unit, and has an $m$-th root in $1+p\Bbb Z_p$ for all $m$ prime to $p$. Thus $S=p\Bbb Z_p$, whose powers are a neighborhood base at zero for the $p$-adic topology. These sets are consequently preserved by the automorphism $\varphi$, so $\varphi$ is continuous.