How many non-equivalent configuration of $2 \times 2 \times 2$ Rubik cubes with $6$ distinct colors?

Question

How many non-equivalent configuration of $2\times2\times2$ Rubik cube with $6$ distinct colors are there?

You can also think this question such a way that how many non-equivalent colorings are there to color $24$ square on a cube whose each faces have $4$ squares.

My try: At first glance, I thought that I can solve it using Polya theorem such that if we call the colors as $a,b,c,d,e,f$, then find $[a^4b^4c^4d^4e^4f^4]$ in the statement constructed by Burdsides' lemma and Polya theorem.

Then, the cumbersome part is to find the statement. Firstly, I assumed that the cube has $24$ symmetries as expected. So, I found the following:

• $\pi_0=x_1^{24}$, the identity permutation

• $\pi_1=\pi_3 =x_4^6$, the $90^\circ$ and $270^\circ$ rotations

• $\pi_2=x_2^{12}$, the $180^\circ$ rotations.

We have three times these rotations, so $3\times (\pi_1+\pi_2+\pi_3)=6x_4^6+3x_2^{12}$:

• For opposite middle points: $6 (x_2^{12})$

• Diagonal $120^\circ$: $x_3^8$

• Diagonal $240^\circ$: $x_3^8$

So, $4 \times 2x_3^8=8x_3^8$ for diagonal rotations.

RESULT: $$\frac{1}{24}[x_1^{24}+9x_2^{12}+8x_3^8 +6x_4^6]$$

$\color{red}{\text{However,}}$ this is not a cube, it is a Rubik cube, so I understood that I have more symmetry than $24$ such that I can also rotate the blocks each consisting of $4$ subcubes, or I can rotate two blocks at the same time. Hence, I have more symmetry.

Related links:

• https://www.quora.com/How-can-we-calculate-the-number-of-permutations-of-a-2x2x2-Rubik%E2%80%99s-cube-Can-you-clearly-explain-how

I want help to solve this question using Polya's technique. Can you help me to find all symmetries and writing correct Polya formula?

Thanks in advance!