Is there always at most one full-support extreme symmetric equilibrium? Given a ($n$-player) symmetric game and two equilibriums $s_1,s_2$, is
  it true that if the support (the set of strategies with positive
  probabilities) of $s_1$ is identical to the support of $s_2$ then
  $\frac{s_1+s_2}{2}$ is also a symmetric equilibrium?




By $\frac{s_1+s_2}{2}$ I mean that a strategy $i$ will be selected by a player with a probability which equals the average of the probability of $i$ under $s_1$ and the probability of $i$ under $s_2$.
This seems obvious.  If $\sigma_1,\sigma_2$ are in the joint support of $s_1,s_2$ and $\tau$ is not, then (writing $P$ for the first player's payoff) you need
$$P(\tau,s)\le P(\sigma_1,s)=P(\sigma_2,s)$$ 

where $s$ is a convex combination of the $s_i$.  But if you replace $s$ with $s_1$ or $s_2$, this holds, so it still holds after you average over $s_1$ and $s_2$.

What am I missing?