Here $F_2$ is the free group on two generators $x,y$. I'm interested in examples of finite groups $G$ such that $Aut(F_2)$ acts transitively on the set of surjections $F_2\rightarrow G$. (In particular I'm interested in $G$'s where automorphisms of $F_2$ of *determinant 1* act transitive on surjections $F_2\rightarrow G$)

In this case the conjugacy class of the image of the commutator $xyx^{-1}y^{-1}$ is an invariant of any $Aut(F_2)$-orbit, so a necessary condition for transitivity of the action is that any two commutators of generating pairs of $G$ are conjugate in $G$

If $G$ is simple then one expects that "most" pairs of elements of $G$ will actually generate $G$. Thus, for alternating groups $A_n$ ($n > 5$), where every element is a commutator, one expects the action of $Aut(F_2)$ can't possibly be transitive.

For groups of the form $G = PSL_2(\mathbb{F}_q)$ it's proven in http://www2.math.ou.edu/~dmccullough/research/pdffiles/traces.pdf

that for $q > 11$ almost every element of $\mathbb{F}_q$ appears as the trace of a commutator of a generating pair, and so again $Aut(F_2)$ can't act transitively in this case.

Is this generally true for all finite nonabelian simple groups? More precisely..

Does there exist a nonabelian simple group $G$ on which $Aut(F_2)$ acts transitively on surjections $F_2\rightarrow G$?

What heuristics are there for describing the groups $G$ admitting such a transitive action (by $Aut(F_2)$? by automorphisms of determinant 1?)