There [exists][1] many closed connected hyperbolic 3-manifolds $M$ with trivial symmetry group, and hence [trivial mapping class group][2]. $M$ cannot be homeomorphic to a simplicial complex $\tau$ which admits a vertex-transitive automorphism group $G$ (which must be non-trivial). Then the quotient $M/G = \tau/G$ is a 3-orbifold which by the [orbifold theorem][3] must be hyperbolic. But this implies that the symmetry group was conjugate in the mapping class group to a group of isometries, a contradiction. 

I imagine that this ought to be true for [symmetry-free hyperbolic manifolds][4] in any dimension >2, but I’m not quite sure how to prove that the automorphism group of the triangulation induces non-trivial outer automorphisms of the fundamental group (equivalently non-trivial isometries by Mostow rigidity). 

For the 2-dimensional case, it does appear to be open in general which closed surfaces admit a vertex-transitive triangulation. [This paper][5] states that there are at most four exceptions with $\chi \geq -127$. 


  [1]: https://mathscinet.ams.org/mathscinet-getitem?mr=953960
  [2]: https://www.ams.org/journals/jams/1997-10-01/S0894-0347-97-00206-3/
  [3]: https://en.wikipedia.org/wiki/Orbifold#3-dimensional_orbifolds
  [4]: https://mathscinet.ams.org/mathscinet-getitem?mr=2132171
  [5]: https://Equivelar%20and%20d-Covered%20Triangulations%20of%20Surfaces.%20I