Using the result by Valiant and Vazirani, we know that Unique-3SAT (3SAT with a unique solution) is hard unless NP=RP. Also it is widely believed that the "Unique" version of any NP-complete problem admits the same result comment after Theorem 2.1.
However, there are NP-complete problems whose instances admit only zero or more than one feasible solutions. Consider, NAE-3SAT where in each feasible solution, each clause must have at least one true and at least one false literal. Thus, by definition, the complement (make true to false and false to true for all variables) of any feasible assignment is also feasible. Thus this problem can have zero or more than one solutions.
Is it possible, to find a parsimonious reduction from Unique-3SAT to Unique-NAE-3SAT? How do you preserve the number of solutions in such a reduction? If there is no such reduction, then is not the widely believed statement as mentioned above is false?