I know that there are mathematical descriptions of the Rubik's Cube based on finite group theory. And as far as I know those descriptions deal with just coloured tiles. Suppose however that in addition to the typical colourings the cube has letters (or markings similar to dice) applied to each tile and those letters are orientable. So that a correct solution to the cube not only requires you to match that correct coloured tiles to form a monochromatics face but also requires that every face have its letters in the same orientation.
How does orientation change our group theoretic description of the cube?
I am not certain if there is a better way to describe these cubes. It is a two years work anniversary gift.
Best Answer
There are many ways you could describe such a Rubik's cube in group theory. One of the easiest ways is to divide each centre piece into four equal squares. Then clearly rotating the centre piece is simply a 4-cycle on those squares. The other pieces of the cube (edges and corners) are divided as usual into 2 and 3 faces respectively, which is sufficient all unique description of any state, because the position and orientation (flip or twirl) of each of those pieces already fully determines the state of each of its faces (unlike for the centre pieces).
Now as to the more interesting question of which states (of the centre pieces) are possible, it turns out that there are $\frac12 \cdot 4^6 = 2048$ states, namely all the states where the sum of the parities of the quarter turns on each centre piece is even. This is easy to prove. First note that each quarter turn moves exactly four corner pieces in a $4$-cycle (ignoring orientation), and thus by basic group theory you need an even number of quarter turns to perform any even permutation of the corners, which includes going from a solved state to a solved state (ignoring centre pieces). Thus the sum of the parities of the centre pieces' rotations (in quarter turns) must be even. Also, it is easy to check that we can achieve any such state by first performing the necessary quarter turns to set the centre pieces in the correct orientation (which performs an even permutation on the edges and an even permutation on the corners) and then use only commutators (as described here) to solve all the edges and corners. (If one is clever enough, one can alternatively use commutators to directly turn the centre pieces!)