Consider a Galois extension $\mathbb{A}/\mathbb{Q}$ of algebraic numbers over the rationals and consider a polynomial $f$ in $\mathbb{Q}[x]$ with roots $\alpha_k$ in $\mathbb{A}$. Consider further that all $\alpha_k$ are not elements of $\mathbb{Q}$ (so that they are not fixed by the automorphisms in the Galois group).
Then consider the automorphisms $g$ in $\operatorname{Aut}(\mathbb{A}/\mathbb{Q})$. Since $g$ is an automorphism, $g(\alpha_i)=\alpha_j$ by the properties of field homomorphisms.
Can anything more be said about which roots are the result of applying the automorphism $g$? For example, $g(\alpha_i)\neq \alpha_i$ for some $g$; can it be said when $g(g(\alpha_i))\neq \alpha_i$, $g(g(g(\alpha_i)))\neq \alpha_i$, and so on? In finite fields this is easier as we can compute the order as $|\operatorname{Aut}(\mathbb{F}_{p^{q_1}}/\mathbb{F}_{p^{q_2}})|=q_1/q_2$, but how can we wrestle with this in the infinite case; with finite subgroups?