The mixed strategy $s=(s_1^*,...,s_n^*)$ is a Nash equilibrium if and only $\forall i \in \{1,...,n\}$ $ \forall s_i \in S_i$ $ u_i(s_1^*,...,s_{i-1}^*,s_i,s_{i+1}^*,...,s_n^*) \le u_i(s_1^*,...,s_{i-1}^*,s_i^*,s_{i+1}^*,...,s_n^*)$ where $s_{i,j}$ is the probability that the strategy $\sigma_{i,j}$ is chosen, $ \sigma \in \Sigma_1 \times...\times \Sigma_n$ the strategy set and $u_i$ is the payoff function.
Q: Let's restrict us on a 2x2 payoff matrix . If I determine the Nash equilibrium (algorithmically), I normally set $\mathbb E[u_i(S_{i_2})] = \mathbb E[u_i(S_{i_1})], i=1,2$. This delivers me a probability vector $(p, 1-p)$ for one player (let's say 1) so that if he choses it, no matter which probability $(q,1-q),$ $ q \in [0,1]$ player 2 takes, the expected payoff stays the same.
Generalizing this idea, I don't see why not $\forall i \in \{1,...,n\}$ $ \forall s_i \in S_i$ $ u_i(s_1^*,...,s_{i-1}^*,s_i,s_{i+1}^*,...,s_n^*) = u_i(s_1^*,...,s_{i-1}^*,s_i^*,s_{i+1}^*,...,s_n^*)$ holds?
In short, I am looking for a example where '$ <$'.