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

[Dyer–Formanek–Grossman](https://doi.org/10.1007/BF01304807 "On the linearity of automorphism groups of free groups") 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][2]).

In their Theorem 1(i), [Charney–Crisp][3] 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](https://doi.org/10.2307/2374228 "The Automorphism Groups of the Braid Groups").

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


  [1]: https://link.springer.com/article/10.1007/BF01304807
  [2]: https://arxiv.org/abs/2306.13437 "Commensurations of Aut(F_N) and its Torelli subgroup"
  [3]: https://arxiv.org/abs/math/0408412 "Automorphism groups of some affine and finite type Artin groups"
  [4]: https://www.jstor.org/stable/2374228?origin=crossref&seq=8