Recall that there are $$\frac{n!}{\prod^n_{i = 1}i^{k_i}k_i!}$$ permutations in $S_n$ which have cycle structure $(k_1, \dots, k_n)$, that is to say they have exactly $k_1$ 1-cycles, $k_2$ 2-cycles, ... and $k_n$ n-cycles. The cycle index of $S_m \times S_n$ acting on the set $\{1, \dots, n\} \times \{1, \dots, n\}$ of indices of entries of an $m \times n$ matrices by row permutations and column permutations can be written as

$$Z(S_m \times S_n; s_1, \dots, s_x) := \frac{1}{m!n!} \cdot \sum_{k_1 + 2k_2 + .. mk_m = m \\ l_1 + 2l_2 + .. nl_n=n} \frac{m!n!}{\prod^m_{i = 1}i^{k_i}k_i! \cdot \prod^n_{j = 1}j^{l_j}l_j! } \cdot \prod_{i = 1}^m \prod_{j = 1}^n s_{\mathrm{lcm}(i,j)}^{\gcd(i,j) \cdot k_i \cdot l_j}$$

In the square case, does anyone know how to calculate the cycle index of the extension $(S_n \times S_n) \rtimes C_2$ where $C_2$ acts by transposing? I.e. extending by the permutation which sends $(i,j)$ to $(j,i)$ for all $1 \leq i,j \leq n$.

Equivalently, if $\sigma \in S_n$ has cycle structure $(k_1, \dots, k_n)$ and $\rho$ has cycle structure $(l_1, \dots, l_n)$ then $(\sigma, \rho, 1)$ has cycle index $$z((\sigma, \rho, 1)) := \prod_{i = 1}^n \prod_{j = 1}^n s_{\mathrm{lcm}(i,j)}^{\gcd(i,j) \cdot k_i \cdot l_j}$$ but if $c$ is the generator of the $C_2$ factor then what is the cycle index of $(\sigma, \rho, c)$?