For positive integers $w$, $h$, $s$ consider the equivalence classes of $w\times h$ matrices, with entries on the set $S=\{1,2,3,\ldots,s\}$, identified by arbitrary permutations of the rows and the columns. The goal is to compute the number of those classes.
This is an application of Polya’s enumeration theorem.
Consider the group $G=S_w\times S_h$, where $S_w$ and $S_h$ are the symmetric group of the sets $W=\{1,2,3,\ldots,w\}$ and $H=\{1,2,3,\ldots,h\}$ with $w$ and $h$ elements, respectively. The group $G$ acts on the set $X=W\times H$, which we view as the set of indexes of the entries of the matrices.
Each matrix is a function $f:X\to S$, and element $f\in S^X$.
So, the problem consists of computing the cardinality
\[\left|S^X/G\right|\]
Cycle index polynomial of a group of permutations
For a permutation group $G$ acting on a set $X=\{1,2,3,\ldots,n\}$ define the polynomial
\[Z(G,s_1,s_2,s_3,\ldots,s_n)=|G|^{-1}\sum_{g\in G}\prod_{k=1}^{n}s_k^{c_k(g)}\]
where $c_k(g)$ is the number of cycles of length $k$ in the cycle decomposition of the permutation $g$.
Polya’s enumeration theorem tell us that
\[\left|S^X/G\right|=Z(G,s,s,\ldots,s)=|G|^{-1}\sum_{g\in G}s^{c_k(g)}\]
Cycle index of the Cartesian product
In our case, $G$ is the Cartesian product $S_w\times S_h$. In this paper they prove the following formula
\[Z(G_1\times G_2,s_1,s_2,\ldots,s_{n_1\cdot n_2}) = Z(G_1,s_1,\ldots,s_{n_1})\otimes Z(G_1,s_1,s_2,\ldots,s_{n_2})\]
where the product \(\otimes\) (In the paper they use the symbol text reference mark) of polynomials is defined by:
If
\[f(x_1,x_2,...,x_m)=\sum a_{i_1i_2\ldots i_m}x_1^{i_1}x_2^{i_2}\dotsm x_m^{i_m}\text{ and }g(x_1,x_2,...,x_n)=\sum b_{j_1j_2\ldots j_n}x_1^{j_1}x_2^{j_2}\dotsm x_n^{j_n}\]
then
\[f(x_1,x_2,\ldots,x_m)\otimes g(x_1,x_2,\ldots,x_n)=\sum a_{i_1i_2\ldots i_m}b_{j_1j_2\ldots j_n}\times\prod_{\substack{1\leq r\leq m\\1\leq s\leq n}}(x_{r}^{i_l}\otimes x_{s}^{j_s})\]
where
\[x_{r}^{i_l}\otimes x_{s}^{j_s} = x_{\operatorname{lcm}(r,s)}^{\gcd(r,s)i_rj_s}\]
Cycle index of the symmetric group
For the case of the symmetric group $S_n$ acting on $\{1,2,3,\ldots,n\}$ we have
\[Z(S_n)=\frac{1}{n}\sum_{k=1}^{n}s_kZ(S_{n-k})\]
or explicitly
\[Z(S_n,s_1,s_2,\ldots,s_n)=\sum_{r_1+2r_2+3r_3+\ldots nr_n=n}\frac{s_1^{r_1}s_2^{r_2}\dotsm s_n^{r_n}}{1^{r_1}r_1!2^{r_2}r_2!\dotsm n^{r_n}r_n!}\]
Solution of the problem
Combining the results above we obtain that the number of classes of $w\times h$ matrices on $s$ symbols up to permutations of the rows and columns is
\[N=\frac{1}{w!h!}\sum_{\substack{i\in P(W)\\j\in P(H)}}\frac{w!}{1^{i_1}i_1!2^{i_2}i_2!\dotsm w^{i_w}i_w!}\frac{h!}{1^{j_1}j_1!2^{j_2}j_2!\dotsm h^{j_h}j_h!}s^{\sum_{\substack{a\in i\\b\in j}}\gcd(a,b)}\]
where the outer sums runs over the partitions $i=(i_1,i_2,\ldots,i_w)$ and $j=(j_1,j_2,\ldots,j_h)$ of $w$ and $h$, respectively. This is, $i_1+2i_2+\ldots+wi_w=w$ and $j_1+2j_2+\ldots+hj_h=h$. The inner sums run over the elements $a$ and $b$ of the partitions $i$ and $j$. This is, $a$ takes the value $1$, $i_1$ times, the value $2$ $i_2$ times, and so on and likewise $b$ takes the value $1$ $j_1$ times, the value $2$ $j_2$ times, etc.
See here.