Given a game $G$, we can construct another $G'$, by a positive scaling i.e. $\lambda \in \mathbb{R}_{++}$, s.t. each entry of $A$ is scaled by $\lambda$
Obviously, $G$ and $G'$ have the same nash strategy. We have simply uniformly increased the payoff of every strategy.
What if we flip the sign of every single payoff entry?
Intuitively, this implies a change from each "payoff" to "cost". In the first game without the sign change, the nash strategy maximizes each player's payoff. In the second game with a sign change, the nash strategy minimizes each player's cost. So the nash strategies should be equivalent.
Does this intuition hold for all normal form/matrix games?
What if we restrict our attention to zero sum games or symmetric games??
(After edit, thanks @Robert Israel) If the payoff is $$ \begin{pmatrix} (1,1) & (0,0) \\ (0,0) & (-1,-1) \end{pmatrix} $$ then the only Nash equilibrium is to play strategy 1 for both players, which changes when you change the signs. Maximizing profit may be equivalent to minimizing cost, but it is definitely not equivalent to maximizing cost.
However, for zero-sum games represented by symmetric matrices, something like that may be true.