If $M$ is a finite volume hyperbolic 3manifold, then its isometry group is finite. I believe this is also true for geometrically finite 3manifolds. What is the most general condition on a hyperbolic 3manifold that is known to ensure that its isometry group is finite?

Here is the detailed answer. First, you have to assume that your hyperbolic manifold is complete and has finitely generated fundamental group, otherwise you will get no answer except for the tautological one. Thus, you are dealing with a finitely generated torsion free Kleinian group G in SO(3,1). If such a group is elementary then the normalizer of the group is not discrete, consider for instance the case of an abelian discrete group of isometries. Thus, let us exclude elementary groups G. Then the normalizer is discrete (consider its action on the limit set of G). Now, if G is also geometrically finite, then the normalizer is a finite extension of G. To see this, consider convex core of your hyperbolic manifold and observe that its thick part is compact. In the geometrically infinite case, there is one example you should be aware of: Take finite volume hyperbolic 3manifold which fibers over the circle and take M to be its infinite cyclic cover associated with the fiber. Then the isometry group of M contains infinite cyclic group. Covering Theorem of Thurston and Canary in addition to the positive solution of tameness conjecture by Agol, Calegari and Gabai shows that such M's are the only examples where the isometry group is infinite. Moreover, the isometry group in this case is a finite extension of an infinite cyclic group. (The latter is immediate.) A side note: If you consider higher dimensional hyperbolic manifolds, the situation in geometrically finite case is essentially the same except you have to assume that the limit set is not contained in a round sphere of codimension 2. In geometrically infinite case, there are more examples one has to exclude and there is no even conjectural classification of exceptions. 

