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Following my previous question about the outer automorphism group, I would like to know if the structure of the automorphism group of the absolute Galois group of the rationals is known. Specifically, is it isomorphic to the cyclic group with two elements ?

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By the Neukirch–Uchida theorem , Aut(Gal($\mathbb{\bar{Q}/Q}$))= Gal($\mathbb{\bar{Q}/Q}$).

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    $\begingroup$ Isn't the theorem that $\operatorname{Out}(\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})) = \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$? $\endgroup$
    – anomaly
    Dec 12 '18 at 13:48
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    $\begingroup$ @anomaly I'm not familiar with Neukirch-Uchida, but it often happens for a non-abelian group that all its automorphisms are inner. A well known example is the symmetric group $S_n$, $n\ge5$, whose automorphisms are all inner except in the case $n=6$. $\endgroup$ Dec 13 '18 at 4:51
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    $\begingroup$ Thanks for the clarification @anomaly. $\endgroup$ Dec 13 '18 at 5:18
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    $\begingroup$ @anomaly In fact Out(Gal($\mathbb{\bar{Q}/Q}$))={1}. Gal($\mathbb{\bar{Q}/Q}$) is center-free, Aut(Gal($\mathbb{\bar{Q}/Q}$))=Inn(Gal($\mathbb{\bar{Q}/Q}$))$\cong$Gal($\mathbb{\bar{Q}/Q}$) $\endgroup$
    – Bonbon
    Dec 13 '18 at 7:02
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    $\begingroup$ @anomaly You can see a proof of this result in the book "Cohomology of Number Fields" by Neukirch and others. $\endgroup$
    – Bonbon
    Dec 13 '18 at 7:04

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