Cycle index of $(S_n \times S_n) \rtimes C_2$ acting on matrix indices by row/column permutation and transposition 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 $(\sigma, \rho, c)$?
 A: More generally, the analogue with any number of factors is studied in E.M. Palmer, R.W. Robinson, Enumeration under two representations of the wreath product, Acta Math. 131 (1973) 123–143. A species approach to this problem can be found in 
Ji Li,
Prime graphs and exponential composition of species,
Journal of Combinatorial Theory, Series A,
115 (2008), 1374–1401.
A: The question is about cycle types of elements of the wreath product $G = (S_n \times S_n) \ltimes C_2 \cong S_n \wr C_2$ in its product action on  $\{1,\ldots, n\} \times \{1,\ldots, n\}$. 
The permutation $(\sigma, \rho, c) \in G$ is conjugate, by $(\rho^{-1},\mathrm{id}_{S_n},1)$, to $g = (\rho\sigma, \mathrm{id}_{S_n}, c)$. Suppose 
$$\rho\sigma = (x_1,\ldots, x_a)(y_1,\ldots, y_b) \ldots .$$ 
Let $X = \{x_1,\ldots, x_a\}$ and $Y = \{y_1,\ldots, y_b\}$. Take all $x$ indices modulo $a$ and all $y$ indices modulo $b$. With this convention,  $(x_i,y_j)g = (y_j,x_{i+1})$ and $(y_j,x_i)g = (x_i,y_{j+1})$. Hence 
$$(x_i,y_j)g^2 = (x_{i+1},y_{j+1})$$ 
and so $(x_1,y_1)$ is in a cycle of $g^2$ of length $\mathrm{lcm}(a,b)$. Since $g$ swaps the disjoint sets $X \times Y$ and $Y \times X$, the cycle of $g$ containing $(x_1,y_1)$ has length $2\mathrm{lcm}(a,b)$. Therefore $g$ acts on $X \times Y \cup Y \times X$ with cycle type 
$$\bigl((2\mathrm{lcm}(a,b))^{ab/\mathrm{lcm}(a,b)}\bigr).$$
For the action on $X \times X$ things are slightly fiddlier: if $a$ is even then there are $a/2$ cycles each of length $2a$; if $a$ is odd then the cycle type is
$$\bigl( (2a)^{(a-1)/2}, a \bigr).$$
Putting these together determines the cycle type of $g$ and hence the cycle index of $G$.
