You're right, just wanted to revert my question because when you think 1 second about it the answer is obvious.
So to be a little bit more constructive I want to stress that maybe you are looking for the Nash-equilibrium from the wrong side. I would conjecture that there will be none if you assume that 3 players play the same strategy and 1 player plays another, because you can always tune the 1 player to counter what the others do. So if they go heavy witch, play for defense, where you are less effected by the curses then them. Additionally, the moats draw the same number of cards as the witches and are less expensive. So you should be on advantage.
On the other hand, if 3 players go for defense you can easily beat them by just not attacking.
But there might be a Nash-equilibrium in the regime where each 2 players play the same strategy. Because if 2 attack and 2 defense, if you join the attackers as defender you might end up in a worse position, and if you switch from attack to defense there is not enough attack left that defense is usefull. But of course this is all not really one-dimensional nor linear, but at least there is more freedom with this approach than fixing the 3 vs 1.
Edit2:
For example. if you let P1 and P2 play your Moat/Moat/Witch/Witch, and P3,P4 play 4xMoat, the winchances are 26:26:22:22. If you switch one of them to {M/M/W/W,BM,4xM}, his winchance will drop.
Edit3:
3xMoat seems to behave slightly better than 4xMoat, but the effect seems to be stable. Interestingly, when you go down to 2 Moats, it is no equilibrium anymor because joining the attacking team will improve one of the defenders. But luckily, the 2xMoat is also worse than 3x Moat when played against 2xMMWW+3M, so that does not contradict the equilibrium.
