Let $G$ be a finite, symmetric, congestion game. According to Nash theorem, a (mixed) symmetric equilibrium surely exists. Congestion games also known to admit pure-strategies Nash equilibrium as they are exact potential games.
I was wondering if a symmetric (i.e. one in which all players play a strategy with the same probability) can yield higher social-welfare (sum of payoffs for the players) than any other equilibrium for the game.
The intuition is that if all players are using the same elements, this has to incur higher cost than if they partition across multiple actions, as the payoff per player depends directly on the number of players using the same elements.
Also, is there always a single symmetric equilibrium for congestion games?