Just to put things in perspective, recall that the Hopf Conjecture asks whether $S^2\times S^2$ admits a metric of positive sectional curvature. By the work of Hsiang-Kleiner, it is known that, if $S^2\times S^2$ admits such a metric, then its isometry group cannot contain a circle, and is hence finite.

Q: If $S^2\times S^2$ admits a metric with $sec>0$, what is known about its isometry group $G$?

The only results I know of are:

0) [Edit suggested by Misha] The diagonal antipodal action of $\mathbb Z_2$ on $S^2\times S^2$, i.e., $\pm 1\cdot(x,y)=(\pm x,\pm y)$, cannot be isometric if $S^2\times S^2$ is equipped with a metric of positive curvature. By Weinstein's Thm, an orientation-preserving isometry of an even-dimensional positively curved manifold has a fixed point (and the antipodal map does not). Equivalently, it would induce a positively curved metric on the $2$-fold orientable cover of $\mathbb R P^2\times \mathbb R P^2$, hence on $\mathbb R P^2\times \mathbb R P^2$, but this contradicts Synge's Thm.

1) From Wilking's thesis (Prop 4.2), any simple subgroup of $G$ is either cyclic or isomorphic to a group in a finite list $F_1,\dots,F_k$ of simple groups. (This is actually true for any finitely generated subgroups of isometries of a manifold with $Ric≥0$).

2) From Fang's paper (Thm 1.2), $G$ cannot have a subgroup of sufficiently large odd order (but this lower bound is huge, since it is estimated with Gromov's universal constant for the total Betti number).

Apart from these, are there other known restrictions on what $G$ can be like?