Game theory post-grad reporting.

In my current research I have stumbled upon a problem that seems simple, but managed to block me for weeks already.

Let's say we have $n$-player game with $2$ pure strategies available to each player, and we know for sure that this game has at least one Nash equilibrium in totally mixed strategies. What would be good sufficient conditions for such game to have exactly one equilibrium in totally mixed strategies? Please, take note, I'm talking about *totally* mixed strategies for *all* players, and do not care of situations where one or more players use pure strategies.

On a hunch it feels like a class of $2 \times 2 \times ... \times 2$ games with exactly one Nash equilibrium in totally mixed strategies should be very large. But to my surprise I can't outline even single meaningful subclass of it.

EDIT 1: My current best guess is that at least following should be provable:

If in any two situations that are distinct by action of only one player no player have same payoff, then game can't have more then one equilibrium in totally mixed strategies.

But unfortunately, even for such weak assumption I don't see how.