Make all cells of the square be the same color

Suppose you managed to make all cells black. Rotate board so that left-upper corner is black. Now place in each cell with coordinates $(i,j)$ remainder of $(i-j)$ modulo $3$: $$ \begin{array}{|c|c|c|c|c} \hline \bf0 & 1 & \bf2 & 0 &\ldots\\ \hline 2 & \bf0 & 1 & \bf2 &\ldots\\ \hline \bf1 & 2 & \bf0 & 1 &\ldots \\ \hline 0 & \bf1 & 2 &\bf 0 &\ldots\\ \hline \vdots & \vdots & \vdots & \vdots & \ddots \end{array} $$ Denote number of black $1$'s as $B$. In beginning $B$ is even, and when all cells are black, $B$ is odd. However, if you repaint all cells in $2\times 3$ rectangle, parity of $B$ remains the same, since there is always two $1$'s in rectangle. That's a contradiction.

I almost found a solution, if You mean 2x3 and 3x2 blocks (in another words if we can rotate our block by 90 degrees). I show how to make it for 5x5, then 5x5 blocks generate 200x200 square.

10101
01010
10101
01010
10101

11001
11010
00100
01011
10011

11111
11011
10101
11011
11111

11111
10001
11111
10001
11111

11111
11111
11111
11111
11111

Try to resolve my turns in each phase. Hint: I only used double turns of 2x3/3x2 inverses. One of the main facts is that when we apply two inverses, the intersection of these inverses will not be inversed. I used double inverses with keeping 2x2 squares and changing other four points which are not in the intersection.

Edit:
As it was told in comments, there is a trouble that color in adjacent blocks eill be inversed. May be we can bypass this..