16.75 (2006)

Yes, it can: in (D. J. Collins, Proc. London Math. Soc. (3), 36 (1978), no. 3, 480–493) it was proved that for integers $r, s$ such that $(r, s) = 1$, $|r| \neq |s|$, and $r - s$ is even, the group $G = \langle a, b \mid a^{-1}b^r a = b^s \rangle$ is isomorphic to $\text{Aut}(\text{Aut}(G))$ (V. A. Churkin, Letter of 5 April 2006).