For the case of coloring the four vertices of a tetrahedron, given by the set $\{1,2,3,4\},$ by $n=2$ colors; have $2^4= 16$ colorings.
The symmetries for rotation, of tetrahedron ($A_4$), are given by the following:
- Identity rotation: $e,$
- Rotations about any vertex, in both clockwise and counter-clockwise directions: $$(234), (432), (123), (312), (134), (413), (124), (412),$$
- Rotations about the midpoints of the opposite edges: $$(12)(34), (13)(24), (14)(23).$$
The set $X_{n=2}$ is divided into five equivalence classes (orbits) based on the mappings induced by action of group elements on the set $X_{n=2}.$
The orbits are of size: $1,4,6,4,1.$
Want to know if the number of colors are changed to $n=3,$ then what are the sizes of orbits, without making detailed tables for the same.
Kindly vet:
Let, the $n=3$ colors are denoted by $a,b,c$ respectively.
The orbits of size $1$ have the elements as :
$aaaa, bbbb, cccc.$
The left elements of the set $X_{n=3}$ are $3^4-3= 78.$
The elements of the set: $X_{n=3}$ of the form $2a-2b, 2a-2c, 2b-2c,$ are given by first choosing two colors, out of three. Next filling two places (out of $4$), from the first chosen one. This gives: $3C2\cdot 4C2\cdot 2= 36.$ Each of these elements , $x,$ has the Stabilizer $|\text{Stab}(x)|=2.$
The rest of the elements are : $78-36= 42.$
These have two orbits of the form: a) $1a-3b, 1a-3c, 1b-3c :$ The number of such elements are: $4C3\cdot 3= 12.$
b) $3a-1b, 3a-1c, 3b-1c :$ The number of such elements are: $12.$
Still $42-24= 18$ elements are left?
Edit #1:
The case of $2a-2b, 2a-2c, 2b-2c,$ gets only $18$ colorings. Say, first choose a particular color out of $3,$ then select two places out of $4$ to fill in by that color. This gives: $3.4C2= 3.6= 18$ ways, but the second color can be any of the two left. This means need be multiplied by $2C1=2$ ways. Then, get $36$ rather than $18$ ways, or colorings. Say, if the first color (first to fill in two places) chosen is $a,$ then : $aaxx, axxa, axax, xxaa, xaax, xaxa.$ And, there are three ways to choose the first color. So, $18$ ways. Then, there are two ways to choose the second color. But, there is symmetry in : $aaxx, axxa, axax, xxaa, xaax, xaxa,$ as only two colors are involved. Hence, a total of $18$ distinct colorings.