Let $M$ be a connected, boundaryless manifold. A tiling of $M$ is a regular cellulation of $M$, and a tiling is regular (i.e. symmetric) if the group of self-homeomorphisms of $M$ that map cells to cells acts transitively on the flags of the cellulation. (Equivalently, the tiling is regular if the corresponding abstract polytope is regular.) Can we always find a regular tiling of $M$?
For example, the surface of a cube represents a regular tiling of the sphere $S^2$, and we can identify its opposite cells to get a regular tiling of $\Bbb{RP}^2$. These examples generalize to higher dimensions (e.g. using a regular 4-polytope to tile $S^3$) and arbitrary quotients of the polytope's symmetry group.
Another construction I've found is as follows: start with a self-dual regular 4-polytope like the 24-cell, inscribe in each (octahedral) cell its dual polyhedron (a cube), enlarge the inscribed duals slightly and truncate the resulting overlaps, joining the intersecting faces (squares truncated to octagons) along the truncation edges. These faces form a regular 2D tiling of a surface with a large genus. I think this procedure also generalizes to higher dimensions.
But is there a regular tiling of every 2D surface? Can we get every spherical 3-manifold? I'm interested in any necessary or sufficient conditions on $M$ for the existence of a regular tiling.