For even ordered Latin squares, we can create squares in which every cell participates in a $2\times 2$ subsquare by using a simple circulant as for $n=6$ below:
0 1 2 3 4 5 1 2 3 4 5 0 2 3 4 5 0 1 3 4 5 0 1 2 4 5 0 1 2 3 5 0 1 2 3 4
Or
0 1 2 3 4 5 1 0 3 2 5 4 2 3 4 5 0 1 3 2 5 4 1 0 4 5 0 1 2 3 5 4 1 0 3 2
(I think this is just a permutation of the one above, please tell me if I am correct or wrong, and how.)
In this square it appears that no cell participates in more than one subsquare. What is the relationship between this square and say, the following
0 1 2 3 4 5 1 0 3 4 5 2 2 4 0 5 1 3 3 5 1 2 0 4 4 2 5 1 3 0 5 3 4 0 2 1
that allows the latter to have cells participating in multiple subsquares and not the first? Does the latter's property or the former's lack of it thereof have something to do with the relation between the circulant's form an the even order, or just one or the other?
Also, do squares with some cells participating in multiple $2\times 2$ subsquares, and all cells participating in at least one $2\times 2$ subsquare exist for higher even order(I am aware than this is possible for $2^n$ as per $\mathbb{Z}/2^n$, but what about for $n\neq 2^n, n= 0mod2$) and so so, for what format?
Is it possible to have squares for at least certain even order with all cells participating in at least two $2 \times 2$ subsquares?