If I have chased through the literature correctly, I think the answer to your question is "yes". Specifically:

[Dyer--Formanek--Grossman][1] showed that $\mathrm{Aut}^+(F_2)$ 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 $\mathrm{Out}(B_4/Z_4)\cong\mathbb{Z}/2\mathbb{Z}$. (Note that they denote $B_4$ as the Artin group $A(A_3)$.) They attribute the theorem to a [1981 paper of Dyer--Grossman][4].

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


  [1]: https://link.springer.com/article/10.1007/BF01304807
  [2]: https://arxiv.org/abs/2306.13437
  [3]: https://arxiv.org/abs/math/0408412
  [4]: https://www.jstor.org/stable/2374228?origin=crossref&seq=8