Consider a 2-player symmetric game given by a payoff matrix $A\in [0,1]^{n,n}$ for the row player (i.e. the column player matrix is $A^t$).
Define the social welfare as the sum of payoffs for both players, i.e. $$SW(i,j)=A(i,j)+A(j,i)$$
Define the social-welfare of a (possibly mixed) equilibrium in a straight forward manner: $$SW(s_1,s_2) = \sum _{i\in [n]}\sum_{j\in [m]}SW(i,j)\Pr_{s_1}(i)\Pr_{s_2}(j)$$
Is it true that for a symmetric equilibrium $s$ (one has to exist from Nash theorm) and a asymmetric equilibrium $a$, $SW(s)\leq SW(a)$?
For example, consider the following simple game:
$A= \left( \begin{array}{ccc} 1/3 & 2/3 \\ 1/3 & 1/6 \\ \end{array} \right) $
And the column player profit, given by $A^t$ is:
$A^t= \left( \begin{array}{ccc} 1/3 & 1/3 \\ 2/3 & 1/6 \\ \end{array} \right) $
There exist a (pure-strategies) symmetric equilibrium where both players play strategy 1. The social welfare of this equilibrium is 2/3.
A (pure-strategies) asymmetric equilibrium exist as well, where some player (say the rows player) plays strategy 1 while the other plays 2, and this gives a social-welfare of 1.
EDIT: Stef's answer made me think if the statement failed because the strategies in the symmetric equilibrium support were different than the ones in the asymmetric equilibrium. I've created a new question to discuss this possibility here.
No, it is not true. Consider the game with payoff matrices $$A=\begin{pmatrix}0&0&1/5\\0&1&0\\1/5&0&0\\\end{pmatrix} \quad \text{ vs } \quad B=A^T=\begin{pmatrix}0&0&1/5\\0&1&0\\1/5&0&0\\\end{pmatrix}$$ Then $e^2 \text{ vs } e^2$ is a symmetric Nash equilibrium with payoffs $(1,1)$ but $e^1 \text{ vs } e^3$ is an asymmetric Nash equilibrium with payoffs $(1/5,1/5)$.