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Feb 14, 2018 at 14:51 comment added Thomas Rot Thank you for this very interesting class of examples. This is very surprising to me, which probably shows my ignorance of this topic.
Feb 13, 2018 at 21:36 comment added Igor Belegradek There are certainly hyperbolic $3$-manifolds with trivial isometry group, and in fact, for each $n>1$ every finite group is the isometry group of hyperbolic $n$-manifold by a theorem of M.Belolipetsky and A.Lubotzky, see arxiv.org/pdf/math/0406607.pdf. The 3d case is due to S.Kojima.
Feb 13, 2018 at 19:27 comment added Ryan Budney I don't believe there is a proof, but I suspect it's widely believed that "most" hyperbolic 3-manifolds have trivial symmetry group. Exactly what definition of "most" one uses could perhaps change the answer. But if you look at the ratio of hyperbolic 3-manifolds with symmetry to ones without, given a bound on the volume (as the volume goes to infinity) this should be zero. If it doesn't take long I'll look through SnapPy's census. . .
Feb 13, 2018 at 19:10 history answered Allen Hatcher CC BY-SA 3.0