Taisong, in your setup you assumed that your morphism maps you to an integral curve over a field and that in $T$ you have a non-empty open set $U$ over which all fibers of $f$ arre connected. Restricting f over $U$ gives a projectve morphism with connected fibers so the push-forward of the structure sheaf is the structure sheafd.

You have to combine the above conjecture that says that hyperbolic varieties only have subvarieties of general type with the conjecture that says that general type varieties can not have dense rational points. The two together tell you that a hyperbolic variety defined over the rationals must have only finitely many points over a number field $K$. Indeed, if you have infinitely many $K$-rational points, their Zariski closure will be a subvariety which can not be of general type which contradicts the fact that there are no such subvarieties.

Ah! I see. I misunderstood the question and thought that $\sigma$ can be any automorphism of $A$ as a variety, not as a group.