Classifying Galois elements using cardinalities of their orbits

Question

Let $g \in G_\mathbb{Q}$, the absolute Galois group of $\mathbb{Q}$. Assume $g$ is not a complex conjugation.

We can look at the orbit of any $q \in \bar{\mathbb{Q}}$ under the action of $g$: $O_g(q) = \{g^n(q) : n \in \mathbb{Z}\}$.

If $O_g(q)$ is a finite set of cardinality $k$, we can say that $q$ is a periodic point of $g$ of period $k$. Otherwise, $p$ is an aperiodic point of $g$.

We can count the number $N_g(k)$ of orbits of each cardinality $k = 1, 2, \dots $. If all these numbers are finite, we can even put them together into some generating function $\nu_g (s)$ attached to $g$ and try to understand its properties.

Can the conjugacy classes of elements of $G_\mathbb{Q}$ be distinguished or classified using the data $\{N_g(k) : k = 1, 2, \dots, \infty\}$, their orbit signatures as it were?

If there are no periodic points at all, or on the other extreme, if there are infinitely many periodic orbits for all possible periods, then the question won't make sense as stated, but could it then be salvaged, say using densities in place of cardinalities or some other normalization device?

Answer 1

First, I claim, there are no "aperiodic" points at all. Indeed, if $q \in \overline{\mathbb{Q}}$, then necessarily $q \in K$ for some finite (wlog normal) extension $K/\mathbb{Q}$. So the open subgroup $G_K \subseteq G_{\mathbb{Q}}$ stabilizes $q$, so the $G_{\mathbb{Q}}$-orbit of $q$ is in fact just a $G_{K/\mathbb{Q}}$-orbit, and hence finite. In particular, some power $g^n$ of $g$ ($n>0$) acts trivially on $q$, and hence $O_g(q)$ is finite. So there are no "aperiodic" points.

Next, I claim, for each $k$, $N_g(k)$ is either $0$ or infinite. Indeed, suppose $N_g(k)>0$. Then there is some $\alpha \in \overline{\mathbb{Q}}$, such that the orbit of $\alpha$ under $g$ has exactly $k$ elements. Then also $\alpha + x$ for any $x \in \mathbb{Z}$ has the same property, i.e., $N_g(x)$ is infinite.

This said, I also don't see a way to rescue the question by using any kind of densities.