Let a permutation matrix, when multiplied on the left, shift the rows down, moving the bottom row to the top. If n is even, we'll negate the original bottom row, so that the determinant of the permutation matrix is 1. Multiply on the right, and you shift the columns left, moving the leftmost to the right and negating it if n is even.
| 0 | 0 | 0 | -1 |
| 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 0 |
I could put -1 in one of the other three positions, but wherever it is, the column negated (before its shift) is one more (by index) than the row negated before its shift. So we could negate row 1 or column 2, then shift down or to the left.
If M is in the center, then multiplying on the left or the right is the same. start with x and y, where y is down and to the right of x. Assume the row containing x is not negated, and the column containing y is not negated. When x shifts down, it is the same as y shifting left; hence x = y. If x and y are on the main diagonal, then x is negated iff y is negated, so there is no trouble. When off diagonal, negate the row of y, and if this also negates the column of y, push this down and to the right, so you are negating the row below y and the column to the right of y. With n at least 3, there is no trouble. Thus the main diagonal is constant, and every line parallel to the main diagonal is also constant.
When n = 2, the matrix has the form [a,b|-b,a]. The product of two such matrices has the same form, and all these matrices commute. This is an unusual case - an abelian group living inside a noncommutative ring. If we go beyond U, we might want to commute with some other matrix [s,0|0,t], where s and t are different. (No trouble here, unless F = Z2.) This means bs = bt, and if s-t can be canceled, then b = 0. The central matrices are multiples of the identity matrix I, as one would expect. Even in Z2, the center is I (you can check this by hand). So let's move on and assume n > 2.
Swap rows 1 and 2, or columns 1 and 2, negating as necessary; and entries 1,3 and 2,3 must be opposite, and 1,4 and 2,4 are opposite, and 3,1 and 3,2 are opposite, and so on. Do this across the board, and M has a constant diagonal in the foreground, against a checkerboard background of + or - x.
| c | x | -x |
| -x | c | x |
| x | -x | c |
Apply a matrix, drawn from U, that looks like the identity, but has -1 in the first two positions. This scales the first two rows, or the first two columns, of M. The result must be the same. Look in position 1,3 and x must equal -x. This sets x = 0. A matrix can be in the center only if it is constant diagonal.
If we are working in characteristic 2, then step outside of U, to a matrix [s,0|0,t], and scale the first row or first column by s, and the second row or second column by t. This gives sx = x, and if 1-s can be canceled, x = 0. Once again the central matrices are constant diagonal.
If the determinant of M must equal 1, then the center is c times the identity matrix, where cn = 1. The 3 by 3 matrices mod 7 have a center of I, 2I, and 4I.
The center of O (the orthonormal matrices), for n > 2, is I, or ±I if n is even.