It is an amusing coincidence (at least it appears to be a coincidence to me) that any completion of the field $\mathbb{Q}$ has trivial automorphism group as an *abstract field*, i.e. when ignoring the topology.

For $\mathbb{R}$, this is obtained as follows: the condition $x\ge 0$ is equivalent to $\exists y:y^2=x$, hence the ordering on $\mathbb{R}$ is preserved by any field automorphism. Since such an automorphism is the identity on $\mathbb{Q}$, it must be the identity anywhere.

For $\mathbb{Q}_p$, one can look at this very short paper. The trick is as follows: $x\in\mathbb{Q}_p^{\times }$ is a unit in $\mathbb{Z}_p$ if and only if it has an $m$-th root in $\mathbb{Q}_p$ for all $m$ prime to $p(p-1)$. Hence $\mathbb{Z}_p^{\times }$ is set-wise preserved by any field automorphism of $\mathbb{Q}_p$, and it is easy to deduce that such an automorphism must be the identity.

Now if we go on to finite extensions, something dramatic happens in the archimedean case: since $\mathbb{C}$ is algebraically closed, its automorphism group is huge.

So, two questions:

(a) Is there any conceptual explanation why $\operatorname{Aut}_{fields }(K)$ is trivial for any completion $K$ of $\mathbb{Q}$?

(b) What happens for finite extensions of $\mathbb{Q}_p$?