I am trying to find a way to express this combinatorial question about the permutation gruop. I am considering a $d$-dimensional lattice $\mathcal{L}$ with $L^d$ sites and periodic boundary conditions, in other words $\mathcal{L} = \mathbb{Z}_L^d$. The elemenst of $\mathcal{L}$ can be expressed as tuples $(i_1,\ldots, i_d)$ with $i_k \in \{1,\ldots, L\}$.
Now, I consider the set of permutations $\sigma \in S_{L^d}$ mapping the elements of the lattice onto themselves. Given a permutation $\sigma$, it has a dimer if for some $k$, it maps $$ \begin{cases} \sigma( i_1,\ldots, i_k, \ldots, i_d) = (j_1,\ldots, j_k, \ldots, j_d) \;, \\ \sigma( i_1,\ldots, i_k + 1, \ldots, i_d) = (j_1,\ldots, j_k + 1, \ldots, j_d) \;, \end{cases}$$ (where the sum are meant $\mod L$. Practically, it maps nearest neighbours in the original lattice onto nearest neighbours in the new one.
I am interesting in computing $N(L,n)$ defined as the number of permutations $\sigma \in \mathcal{S}_{L^d}$ which have exactly $n$ dimers.
The maximal number of dimers $n = d L^d$ corresponds to permutations which are rigid translations and therefore $N(L, d L^d) = L^d$. In general $N(L,n)$ will be always divisible by $L^d$ because starting from $\sigma$, we can build $L^d$ permutations with the same number of dimers applying rigid tranlations.
For $d=1$, I have found a way to express $N(L, n)$ as $$ N(L,d) = L \binom{L}{n} a_{L-n} $$ where $a_k$ are the sequence A000757. Even for $d=2$, I have not found any formula or reference to look at this problem. The problem is that, it is computationally hard to enumerate $N(L,n)$ by brute force already from $L = 4$, since $4^2!$ is huge.