6*6 grid, with 1 person in 1 grid, in 1s each person moves to a grid adjacent to him, find the probability that after 1s all the grids have one person I just encounter a question, when this is 5*5, it is impossible. (Chessboard)
Now I'm curious about what will happen if n is even, for example, 6*6.

6*6 grid, with 1 person in 1 grid, in 1s each person moves to a grid adjacent to him, with each possible movement equally likely, find the probability that after 1s all the grids have one person

I've attempted to solve some simple cases, but can not find any generalized method to count it.
 A: Partial answer: A set of movements having the desired property exists on an $\ n\times n\ $ grid if and only if $\ n\ $ is even. The movements must follow the edges of a set of disjoint $\text{"cycles"}^*$ of the graph of the grid.  Thus a set of such movements is possible if and only there exists such a set of disjoint cycles, the union of whose vertices contains all the vertices of the graph. Every cycle in the graph of a grid must contain an even number of vertices, so the total number of vertices must be even, which is only the case when $\ n\ $ is even.
Conversely, if $\ n=2r\ $ is even, then the union of the vertices of the disjoint $2$-cycles
$$
(2i-1,j)\rightarrow (2i,j)\rightarrow (2i-1,j)
$$
for $\ i=1,2,\dots, r, j=1,2,\dots,n\ $ includes all the vertices of the grid's graph.
The probability that a completely random set of movements will follow any given such set of cycles is
$$
\left(\frac{1}{2}\right)^4\left(\frac{1}{3}\right)^{4n-8}\left(\frac{1}{4}\right)^{(n-2)^2}\ ,
$$
so the probability of such a random set of movements having the desired property is
$$
\left(\frac{1}{2}\right)^4\left(\frac{1}{3}\right)^{4n-8}\left(\frac{1}{4}\right)^{(n-2)^2}C_n\ ,
$$
where $\ C_n\ $ is the number of ways that the graph of the grid can be decomposed into a set disjoint cycles whose vertices include every vertex of the grid's graph. I have no idea, however, whether there exists a convenient formula or algorithm for computing the value of $\ C_n\ $.
$\,^*$ By using the term "cycle" here, I mean to include walks of the form $\ \left(x_1,y_1\right)\rightarrow\left(x_2,y_2\right)\rightarrow\left(x_1,y_1\right)\ $ by following which two adjacent people would exchange places.  Such walks do not satisfy the usual definition of "cycle" in an undirected graph, because the edge $\ \left\{\left(x_1,y_1\right), \left(x_2,y_2\right)\right\}\ $ is repeated.
