Finding equivalence classes under permutation symmetry

Question

If we write a 9 digit binary number as a $3\times3$ matrix, for example $101110101$ would be written as, $$\begin{bmatrix}1 & 0 & 1\\\ 1 & 1 & 0\\\ 1 & 0 & 1 \end{bmatrix}$$ We can take genral representation for any 9 digit binary string as follows, $$I = \begin{bmatrix}I_{00} & I_{01} & I_{02}\\\ I_{10} & I_{11} & I_{12}\\\ I_{20} & I_{21} & I_{22} \end{bmatrix}$$ Now, If I put every number represented by permutations of the indices into the same set, then how many such sets would I have?
To give an example, we apply permutation $\pi_{1,0,2}$ on $I$ $$\pi_{1,0,2}(I) = \begin{bmatrix}I_{11} & I_{10} & I_{12}\\\ I_{01} & I_{00} & I_{02}\\\ I_{21} & I_{20} & I_{22} \end{bmatrix}$$ A more specific example would be $111100001$ becomes $001010111$ by permutation (1,0,2). So we have to essentially divide these $2^9$ numbers into $x$ sets, what is $x$ ?

Answer 1

As pointed out in a comment, you can count these equivalence classes using Burnside’s lemma. There are $6$ symmetry elements. The identity leaves $2^9$ strings invariant. Each transposition leaves one entry of the matrix in place and swaps the other $8$ in pairs, and thus leaves $2^5$ strings invariant. Each $3$-cycle permutes the entries of the matrix in three $3$-cycles and thus leaves $2^3$ strings invariant. Thus the number of orbits under these permutations is

$$ \frac{2^9+3\cdot2^5+2\cdot2^3}{6}=104\;. $$