Non-negative deviations from Nash Equilibrium I know that in a Nash Equilibrium, no player can profitably deviate from the equilibrium strategy assuming that the strategies of the other players remain the same. My question is, what if a player can deviate to a different strategy that generates an equivalent payoff? Would this no longer be a Nash Equilibrium?
It depends on whether the other players' strategies in the initial equilibrium are still best responses to the deviator's new strategy. For example, $(T,L)$ is the unique NE in the game below. Player 1 (the row player) can deviate from $T$ to $B$ without hurting his own payoff, but $(B,L)$ is not a NE, since $L$ is not a best response to $B$ in this game. 

\begin{array}{|c|c|c|}\hline
&amp;L&amp;R\\\hline
T&amp;1,1&amp;2,0\\\hline
B&amp;1,0&amp;0,1\\\hline
\end{array} 

Of course you can easily construct a game where such a unilateral deviation still leads to a NE. As the game below shows, player 1's deviation from $T$ to $B$ is still without loss of profit, but it changes the NE from $(T,L)$ to $(B,L)$. There are also infinitely many other possible deviations for player 1 if you consider deviations in mixed strategies.  

\begin{array}{|c|c|c|}\hline
&amp;L&amp;R\\\hline
T&amp;1,1&amp;2,0\\\hline
B&amp;1,2&amp;0,1\\\hline
\end{array}