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The automorphism groups $\mathrm{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q)$ of algebraic closures of finite fields $\mathbb{F}_q$ and the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ share a wonderful property: they all act in finite orbits on the underlying field ($\overline{\mathbb{F}_q}$, respectively $\overline{\mathbb{Q}}$).

Question: are there other algebraically closed fields which also have this property ? (And if so, are those classified / characterized ?)

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    $\begingroup$ Your question is confusingly worded because it's unclear what “this property” refers to. The absolute Galois group of any perfect field $K$ acts on the algebraic closure $\bar K$ of $K$ with finite orbits, since orbits are sets of conjugates and every element of $\bar K$ has finitely many algebraic conjugates over $K$. Now I think your question actually refers to absolute automorphism groups and not absolute Galois groups, but the way it's asked it's pretty much impossible to tell. $\endgroup$
    – Gro-Tsen
    Commented Jun 23, 2023 at 15:23

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The answer is no. Let $K$ be an algebraically closed field, and let $k$ be the algebraic closure of the prime field contained in $K$. Pick a transcendence basis $B$ of $K/k$, which is nonempty if $k\neq K$. Then $K$ is the algebraic closure of $k(B)$, and any automorphism of $k(B)$ extends to an automorphism of $K$. Now it is enough to show that $k(B)$ has automorphism groups with infinite orbits - for instance, note that for any $x\in B$ and $a\in k$, there is a (unique) automorphism of this field mapping $x$ to $x+a$ and fixing all other elements of $B$.

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    $\begingroup$ In fact, all transcendental elements form one huge orbit: if $a,b$ are transcendental, there is an isomorphism $k(a)\simeq k(b)$ that maps $a$ to $b$, and this extends to an automorphism of $K$. $\endgroup$ Commented Jun 23, 2023 at 13:27
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    $\begingroup$ So in general, by Emil's remark, all orbits are finite except possibly one! $\endgroup$ Commented Jun 23, 2023 at 14:59

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