Let $Aut(\bar{Q})$ be the automorphism group on the field of algebraic complex numbers. The order of an element $f \in Aut(\bar{Q})$ is the least natural number $n$ (if there exists one) such that $(f)^{n}$ is identity. What are the possible orders of elements of $Aut(\bar{Q})$?
This amounts to the ArtinSchreier theorem, which has come up several times already on MO (c.f. Examples of algebraic closures of finite index): if $K/F$ is a field extension with $K$ algebraically closed and $[K:F] < \infty$, then $[K:F] = 1$ or $2$, and in the latter case, $F$ is realclosed. Thus the answer here is that $n$ can be $1$, $2$ or $\infty$, and all possibilities occur: the field of real algebraic numbers gives an index $2$ subfield of $\overline{\mathbb{Q}}$. (Also, just to be sure, there are elements of infinite order! E.g., if not then every element would have order $1$ or $2$, so the absolute Galois group would be abelian, and thus every finite Galois group over $\mathbb{Q}$ would be abelian, and this is certainly not the case.) 

