Let $K$ be a Galois extension of $\mathbb{Q}$. Is it possible for non-trivial elements in $\text{Gal}(K/\mathbb{Q})$ to be continuous everywhere on $K$? Yes -- for instance, if $K$ is not a real number field, then complex conjugation is a non-trivial automorphism that is continuous.
But if $K$ is real, then non-trivial automorphisms are never continuous.
Theorem: Let $S\subseteq\mathbb{R}$ be a set of real numbers that contains $\mathbb{Q}$. Let $f:S\rightarrow\mathbb{R}$ be a function that fixes every rational number. Then if $f$ is not the identity map, it is discontinuous somewhere in $S$.
If $f$ is not the identity, there is some $s\in S$ such that $f(s)\neq s$. Let $s'=f(s)$, and let $D=|s-s'|>0$.
For every $\delta>0$, there is some rational number $q$ such that $|q-s|<\delta$. $S$ contains $\mathbb{Q}$, so $q\in S$.
$|f(q)-f(s)|=|q-s'|=|(q-s)-(s'-s)|\geq |s'-s|-|q-s|>D-\delta$.
So for every $0<\delta<D/2$, there is some rational number $q$ such that $|q-s|<\delta$ and yet $|f(q)-f(s)|>D/2$. However close you get to $s$ on the number line, you will find a rational number that is at a distance greater than $D/2$ from $f(s)$. $\blacksquare$
So if $K$ is a real Galois extension, then non-trivial automorphisms of $K$ are not continuous everywhere.
Question: In general, is complex conjugation the only possible continuous automorphism of a Galois extension of $\mathbb{Q}$?
Update: Here is a special case:
Theorem: Let $K$ be a non-real Galois extension of $\mathbb{Q}$, let $\mathcal{G}=\text{Gal}(K/\mathbb{Q})$, and let $t\in\mathcal{G}$ be complex conjugation. Suppose $t$ is central in $\mathcal{G}$. Then $t$ is the only non-trivial element of $\mathcal{G}$ that is continuous as a function on $K$.
Proof:
Let $x\in K\cap\mathbb{R}$. Then $x^t=x$.
Let $\sigma\in\mathcal{G}$. Then $(x^\sigma)^t=x^{\sigma t}=x^{t\sigma}=(x^t)^\sigma=x^\sigma$.
Thus, $x^\sigma\in K\cap\mathbb{R}$.
So $K\cap\mathbb{R}$ is closed under the action of $\mathcal{G}$.
Recall that we have shown that a real number field has no continuous non-trivial automorphisms. Thus, for every continuous $\sigma\in\mathcal{G}$, the restriction of $\sigma$ to $K\cap\mathbb{R}$ must be the identity map, which means $\sigma\in\text{Gal}(K/K\cap\mathbb{R})=\langle t\rangle$.
Thus, $t$ is the only non-trivial element of $\mathcal{G}$ that is continuous as a function on $K$. $\blacksquare$
Corollary: If $K$ is an abelian Galois extension of $\mathbb{Q}$, then the only continuous non-trivial element of $\text{Gal}(K/\mathbb{Q})$ is complex conjugation.