This post was inspired by this webpage of mathematical challenge due to Mickaël Launay (French).
Let $G_n$ be the subgroup of $S_{n^2}$ generated by the red arrow permutations as for the following picture with $n = 5$:
(Source: http://www.micmaths.com/defis/defi_07.html)
We can call $G_n$ the n-th Rubik Square permutation group.
Question: Is $G_n$ a strict subgroup of $S_{n^2}$?
Let us look at the commutator of the top row and leftmost column operations. We easily see that the move sequence $\leftarrow\uparrow\rightarrow\downarrow$ is a 3-cycle involving the entries at positions $(1,1), (1,2), (2,1)$. Similarly for any other pair of rows and columns. Using appropriate powers of the elementary moves we get 3-cycles on any triple of positions $(i,j), (i,k), (\ell,j)$. A lot hinges on what we can say about the subgroup generated by such 3-cycles.
I claim that such 3-cycles generate the alternating group $A_{n^2}$. It suffices to prove that we get all the 3-cycles. The key observation is that any two 3-cycles $\in A_4$ generate all of $A_4$ unless they share a fixed point. This is because the generated subgroup is doubly transitive, and $A_4$ has no proper doubly transitive subgroups. What this means is that any pair of 3-cycles with an overlap of two positions gives us any 3-cycle among the four positions in the union.
So let $(i_t,j_t),t=1,2,3,$ be any three positions. The 3-cycles on $(i_1,j_1),(i_2,j_2),(i_2,j_1)$ and $(i_1,j_1),(i_2,j_1),(i_2,j_3)$ thus generate the 3-cycles on $(i_1,j_1),(i_2,j_2),(i_2,j_3)$. Together with the 3-cycles $(i_2,j_2),(i_2,j_3),(i_3,j_3)$ this then generates the desired 3-cycle.
Therefore $A_{n^2}\le G_n$.
If $n$ is odd, then all the generators of $G_n$ are in $A_{n^2}$, so we must have $G_n=A_{n^2}$. But if $n$ is even, then $G_n$ contains also odd permutations and we have $G_n=S_{n^2}$.
I won't say anything about the Cayley graph.