Are symmetric equilibria monotone?
For a simple counterexample let
$$
A = B = \left( \begin{array}{cc}
1 &amp; 0 \\
1 &amp; 0 \end{array} \right).
$$
In this game any strategy pair will constitute a Nash-equilibrium. Let
$$
a^t = \left(\frac{2}{3}, \ \frac{1}{3} \right), b^t = \left(\frac{1}{3}, \ \frac{2}{3} \right).   
$$ 
Then the expected payoffs are larger in the symmetric equilibrium associated with $a$.

If you wish for $B$ to be strictly larger than $A$ you can easily modify this example, simply multiply all of $B$'s elements with a number larger than 1 but smaller than 2.

A counterexample where neither $A$ nor $B$ are singular:
$$
A = \left( \begin{array}{cc}
4 &amp; 0 \\
6 &amp; -4 \end{array} \right), \ 
B = \left( \begin{array}{cc}
4 &amp; 1 \\
6 &amp; 0 \end{array} \right)
$$
$$
a^t = \left(\frac{2}{3}, \ \frac{1}{3} \right), b^t = \left(\frac{1}{3}, \ \frac{2}{3} \right).   
$$
$$
A \cdot a = \left( \begin{array}{c}
\frac{8}{3}  \\
\frac{8}{3}  \end{array} \right) &gt;
\left( \begin{array}{c}
2  \\
2  \end{array} \right) = B \cdot b 
$$