I have two $p\times q$ matrices $A$ and $B$ that belong to the set: $$ \{X\in R^{p\times q}: \sum_{i,j} X_{i,j}^2=1 ; \sum_{i,j} X_{i,j}=0 \} $$
Also, showing the $i$'th column of $A$ by $a_i$, I know that $$ (a_i - a_j) = c_{ij} (b_i-b_j) ; \qquad c_{ij}>0 $$ which means that $(a_i - a_j)$ and $(b_i - b_j)$ are parallel and in the same direction for all $i$ and $j$.
Can I conclude $A=B$ ?
Hint: Imagine all the columns of $A$ and $B$ are multiples of a common vector. Can you come up with two different solutions $A_1,B_1$ and $A_2,B_2$ which both have mean zero of entries and sum of squared entries equal to 1? Further hint: Assume all columns of $A_1$ and $A_2$ are equal except for a certain number of vectors, and similarly for corresponding column vectors of $B_1$ and $B_2$.