I will answer the questions for two-player zero-sum games and then show how to extend them to the general case.
Let player Alice have two strategies, $L,R$ and player Bob two strategies $T,B$. Let the payoff of Alice and Bob be constant, for every strategy profile, Alice receives a payoff of $1$ and Bob a payoff of $-1$. The game is clearly zero-sum and every strategy profile is a Nash equilibrium. Moreover, when Alic deviates from any Nash equilibrium, she still gets the positive payoff $1$.
Now lets consider a family of games parametrized by $\alpha>0$ with the same strategies, but the payoff of Alice is given by $u_A(L,T)=u(L,B)=\alpha$ and $u_A(R,T)=u_A(R,B)=-\alpha$. The payoff of Bob is given by $u_B=0-u_A$. This is clearly a zero-sum game for every $\alpha$ and a Nash equilibrium is any strategy profile in which Alice plays $L$. Her loss from switching to $R$ is $2\alpha$ and that can be made arbitrarily large by increasing $\alpha$.
So for two-player, zero-sum games, the answer to both questions is no. Now we can add any numbers of players with arbitrary finite strategy spaces by letting their payoff be constantly $0$ and their actions not affect the payoff Alice and Bob get. This allows us to extend the examples above to any number of players and show that the answer is no in both cases again.
It should be remarked that even though the examples are somewhat pathological and non-generic, one can construct more complicated generic examples. A special case that is natural for zero-sum games are games where all euilibria are completely mixed, that is where every strategy is played with positive probability. In such an equilibrium, the cost of deviating is always zero.
As a final remark, $n$-player zero sum games are rather pointless. We can always add a dummy player who doesn't effect anyone elses payoff and gets the sum of payoffs of all others times $-1$. So every $n$-player game can be interpreted as a $(n+1)$-player zero-sum game.