The answer is $2(n!+1)$ , attained by and only by grids which are either uniform or are a permutation of a ``Sudoku." Clearly these work.* Necessity hinges on the following:

Claim. No row or column can have both $\ge2$ $R$ s and $\ge2$ $B$ s.
Proof. Suppose otherwise, and WLOG arrange them in a row in the order $R_1B_1R_2B_2$ . We build up the grid as follows:

First, swap the columns with $R_1$ and $B_1$ . To revert, there should exist a second row as shown, where the dashed sections are the same in both rows:
$[asy] label("$R_1$",(-3,1)); label("$B_1$",(-1,1)); label("$R_2$",(1,1)); label("$B_2$",(3,1)); label("$B$",(-3,-1)); label("$R$",(-1,-1)); label("$R$",(1,-1)); label("$B$",(3,-1)); draw((-4,1)--(-3.3,1),dashed); draw((-2.5,1)--(-1.5,1),dashed); draw((-0.5,1)--(0.5,1),dashed); draw((1.5,1)--(2.5,1),dashed); draw((3.5,1)--(4,1),dashed); draw((-4,-1)--(-3.3,-1),dashed); draw((-2.5,-1)--(-1.5,-1),dashed); draw((-0.5,-1)--(0.5,-1),dashed); draw((1.5,-1)--(2.5,-1),dashed); draw((3.5,-1)--(4,-1),dashed); [/asy]$
Next, swap these rows. To revert, it is necessary and sufficient to swap the columns with $R_1$ and $B_1$ , so that the solid sections are the same in both columns:
$[asy] label("$R_1$",(-3,1)); label("$B_1$",(-1,1)); label("$R_2$",(1,1)); label("$B_2$",(3,1)); label("$B$",(-3,-1)); label("$R$",(-1,-1)); label("$R$",(1,-1)); label("$B$",(3,-1)); draw((-4,1)--(-3.3,1),dashed); draw((-2.5,1)--(-1.5,1),dashed); draw((-0.5,1)--(0.5,1),dashed); draw((1.5,1)--(2.5,1),dashed); draw((3.5,1)--(4,1),dashed); draw((-4,-1)--(-3.3,-1),dashed); draw((-2.5,-1)--(-1.5,-1),dashed); draw((-0.5,-1)--(0.5,-1),dashed); draw((1.5,-1)--(2.5,-1),dashed); draw((3.5,-1)--(4,-1),dashed); draw((-3,-4)--(-3,-1.5)); draw((-3,-0.5)--(-3,0.5)); draw((-3,1.5)--(-3,2)); draw((-1,-4)--(-1,-1.5)); draw((-1,-0.5)--(-1,0.5)); draw((-1,1.5)--(-1,2)); [/asy]$
Finally, swap the columns with $R_2$ and $B_2$ . To revert, there should exist a third row*
specifically, at least one -- more accurately, there should be two such row, one for each of the already existing rows -- but I omitted that in my solution as it isn't necessary
as shown, where the dashed sections are the same in both rows (in order to match the first row):
$[asy] label("$R_1$",(-3,1)); label("$B_1$",(-1,1)); label("$R_2$",(1,1)); label("$B_2$",(3,1)); label("$B$",(-3,-1)); label("$R$",(-1,-1)); label("$R$",(1,-1)); label("$B$",(3,-1)); label("$R$",(-3,-3)); label("$B$",(-1,-3)); label("$B$",(1,-3)); label("$R$",(3,-3)); draw((-4,1)--(-3.3,1),dashed); draw((-2.5,1)--(-1.5,1),dashed); draw((-0.5,1)--(0.5,1),dashed); draw((1.5,1)--(2.5,1),dashed); draw((3.5,1)--(4,1),dashed); draw((-4,-1)--(-3.3,-1),dashed); draw((-2.5,-1)--(-1.5,-1),dashed); draw((-0.5,-1)--(0.5,-1),dashed); draw((1.5,-1)--(2.5,-1),dashed); draw((3.5,-1)--(4,-1),dashed); draw((-3,-4)--(-3,-1.5)); draw((-3,-0.5)--(-3,0.5)); draw((-3,1.5)--(-3,2)); draw((-1,-4)--(-1,-1.5)); draw((-1,-0.5)--(-1,0.5)); draw((-1,1.5)--(-1,2)); draw((-4,-3)--(-3.3,-3),dashed); draw((-2.5,-3)--(-1.5,-3),dashed); draw((-0.5,-3)--(0.5,-3),dashed); draw((1.5,-3)--(2.5,-3),dashed); draw((3.5,-3)--(4,-3),dashed); [/asy]$
The $R$ and $B$ in corresponding sections of the solid sections give us a contradiction. $\square$

To finish, clearly there number of $R$ 's in each of the rows and columns is a constant $c$ , and the above claim implies that $c\in\{0,1,n-1,n\}$ , which produces those and only those mentioned at the beginning. $\blacksquare$