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Given a field $k$ let $\Omega(k)$ be the set of algebraic closures of $k$.

$\Omega(k)$ is obviously a groupoid. At each element $\bar{k}$ of $\Omega(k)$ we have its automorphism group over $k$, which is a copy of $Gal(\bar{k}/k)$. Any two copies can be identified via isomorphisms of respective algebraic closures.

Question 1: What is this structure "groupoid fibered in (automorphism) groups" called? Is there a reference for it?

Question 2: Can we see automorphisms of $Gal(\bar{k}/k)$ in a natural way in this set up, e.g., via some fiber functor? (In fact what are the automorphism groups of some well-known Galois groups, for $k = \mathbb{Q}$ or $\mathbb{F}_p$ for example?)

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    $\begingroup$ If I understand your Question 1 properly, you're describing the property that every element has the 'same' automorphism group. Doesn't every connected groupoid have this property? $\endgroup$
    – LSpice
    Commented Apr 10, 2020 at 1:52
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    $\begingroup$ The galois group over a finite field is just Z $\endgroup$
    – Asvin
    Commented Apr 10, 2020 at 2:00
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    $\begingroup$ Ctd: is just the profinite completion of Z so any automorphism is determined by sending 1 to any generator. For Q, maybe we can get some mileage out of the fact that automorphisms have to preserve $\endgroup$
    – Asvin
    Commented Apr 10, 2020 at 2:02
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    $\begingroup$ For Question 2, specifically what the automorphism group of Gal$(\overline{\mathbb{Q}} / \mathbb{Q})$ may be, you may be interested in [Kanno, Tsuneo Automorphisms of the Galois group of the algebraic closure of the rational number field. Kōdai Math. Sem. Rep. 25 (1973), 446–448.] and references therein. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Apr 10, 2020 at 2:33

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