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$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Inn{Inn}\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}$Let $F_2$ be a free group of rank 2. There is a surjection $\Aut(F_2)\rightarrow \GL(2,\mathbb{Z})$ with kernel $\Inn(F_2)$, induced by the abelianization map $F_2\rightarrow\mathbb{Z}^2$. Define $\Aut^+(F_2)$ to be the inverse image of $\SL(2,\mathbb{Z})$. Is every automorphism of $\Aut^+(F_2)$ induced by conjugation by something in $\Aut(F_2)$?

Since there seems to be some confusion in the comments, by an "automorphism of $\Aut^+(F_2)$", I mean an automorphism of the group $\Aut^+(F_2)$, not an automorphism of $F_2$.

This is analogous to the known result that $\Aut(F_2)$ is complete (due to Dyer–Formanek "The automorphism group of a free group is complete").

This is also related to results of Ivanov and Farb–Handel: If $\Gamma$ is a finite index subgroup of either a mapping class group of a punctured surface genus $\ge 2$ or any surface of genus $\ge 3$, or the automorphism group of a free group of rank $\ge 4$, then every automorphism of $\Gamma$ is induced by conjugation by something in the ambient group.

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    $\begingroup$ Just to rephrase what is already written in this question. This is equivalent to asking whether any automorphism of $Aut^+(F_2)$ can be extended to an automorphism of $Aut(F_2)$. $\endgroup$
    – HenrikRüping
    Commented Feb 14 at 13:53
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    $\begingroup$ @DanielAsimov, re, I think the question would become nonsensical, or at least trivial, if "automorphism of $\operatorname{Aut}^+(F_2)$" meant "element of $\operatorname{Aut}^+(F_2)$", so surely the only sensible reading is as "automorphism of $G$" in the usual sense, where $G$ happens to be $\operatorname{Aut}^+(F_2)$. $\endgroup$
    – LSpice
    Commented Feb 14 at 18:37
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    $\begingroup$ @DanielAsimov An automorphism of a group is pretty unambiguous IMO? IMO anyone writing "an automorphism of Aut(G)" to mean an automorphism of $G$ would be abusing notation, and the onus would be on them to make this clear. This is not the case in my question. $\endgroup$
    – stupid_question_bot
    Commented Feb 14 at 21:05
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    $\begingroup$ @DanielAsimov, RyanBudney: looking back over previous edits, the first version of the question was entirely unambiguous. $\endgroup$
    – HJRW
    Commented Feb 15 at 10:20
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    $\begingroup$ I think I realized my mistake. There is an elliptic involution of the twice punctured torus which exchanges the two punctures and is central in the mapping class group. So it acts trivially on $Aut(F_2)$. I’ll leave my comment up in case anyone gets confused about the same issue. $\endgroup$
    – Ian Agol
    Commented Feb 16 at 14:09

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$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Out{Out}$If I have chased through the literature correctly, I think the answer to your question is "yes". Specifically:

Dyer–Formanek–Grossman showed that $\Aut^+(F_2)$ is isomorphic to $B_4/Z_4$, the quotient of the 4-strand braid group by its centre. (Bridson–Wade give a geometric proof here).

In their Theorem 1(i), Charney–Crisp state that $\Out(B_4/Z_4)\cong\mathbb{Z}/2\mathbb{Z}$. (Note that they denote $B_4$ as the Artin group $A(A_3)$.) They attribute the theorem to a 1981 paper of Dyer–Grossman.

In summary, $\Out(\Aut^+(F_2))$ has order 2, and hence $\Aut(\Aut^+(F_2))$ must be $\Aut(F_2)$.

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  • $\begingroup$ To add a little detail, a more precise reference to Charney-Crisp's claim is section 4 of the 1981 paper of Dyer-Grossman. $\endgroup$
    – stupid_question_bot
    Commented Feb 17 at 18:29
  • $\begingroup$ Another detail: One should rule out the possibility that the nontrivial coset of $Aut(F_2)/Aut^+(F_2)$ acts as an inner automorphism of $Aut^+(F_2)$. Since $Aut^+$ has trivial center, this would imply that the centralizer of $Aut^+$ inside $Aut$ has order 2, and makes the sequence $1\to Aut^+\to Aut\to C_2\to 1$ a split sequence with trivial action, hence a direct product. However, passing to the quotient by $Inn$, this would imply that $GL_2(\mathbb{Z})$ is a direct product of $SL_2(\mathbb{Z})$ with $C_2$, which is not true. Can you add this to your answer? Then I can accept. Thank you! $\endgroup$
    – stupid_question_bot
    Commented Feb 17 at 18:41

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