<s>[INST] Group of Rotations. 
Attempt:
$$\frac{(2^6 + 6 \cdot 2^3 + 3 \cdot 2^4 + 8 \cdot 2^2 + 6 \cdot 2^3)}{24} = 10$$
 [/INST] $10$ appears correct for distinct cubes allowing rotation mapping. There are:
- $2$ solid-colour cubes $(B^6,R^6)$,
- $2$ cubes with one face different $(B^5R, BR^5)$,
- $2$ (opposite-face and adjacent-face) options for each of the $4{+}2$ colourings $B^4R^2, B^2R^4$
- and $2$ similar options for the evenly-split colouring $B^3R^3$,
which overall gives $1+1+2+2+2+1+1 = 10$
The restriction to rotations (no reflections) makes no difference in this case.
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