If a game has a unique Nash Equilibrium, then does it have a unique Mixed Nash Equilibrium as well, where this MNE is the unique NE?
The game I have in mind is the following (but I am more curious about any game in general).
$$ \begin{array}{|c|c|c|} \hline & L & R \\ \hline U & 5,6 & 3,1 \\ \hline D & 1,4 & 2,3 \\ \hline \end{array}$$
Sorry about the really poor formatting btw.
Anyway, it seems intuitive that the outcome would be (U,L) as this is more beneficial than mixing over the other possibilities. I get there because if you suppose the column player mixes in general (playing L with probability q), then $5q + 3(1-q) = q + 2(1-q) \Rightarrow q = \frac{-1}{3}$ which is impossible and so the column player doesn't mix.
If a game has a unique Nash Equilibrium, then it can be Pure or Mixed Nash Equilibrium, whichever exists. In your case, the unique Nash eq is in pure strategies. (Note: By the way, a Pure Nash Equilibrium is itself a Mixed Nash Equilibrium. It is an extreme case in which players assign probability 1 to one strategy and probability 0 to all of their other strategies.)