Here is a counterexample (from p.76 of my thesis):
$$ A = A^T = \begin{bmatrix} 0 & 3 & 2 \\ 3 & 0 & 2 \\ 2 & 2 & 3 \end{bmatrix}. $$
Labeling the strategies in order as $a$, $b$, and $c$, there are asymmetric Nash equilibria $(a,b)$ and $(b,a)$ with support $\{a,b\}$, but there is no symmetric Nash equilibrium supported on this set. If the players play the same mixed strategy supported on this set then at least half of the time they will receive zero payoff, so their expected payoff can be at most $\frac{3}{2}$. But then they are better off deviating to $c$.