There exists many closed connected hyperbolic 3-manifolds $M$ with trivial symmetry group, and hence trivial mapping class group. $M$ 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 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 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).