How hard is it to count the number of local optima for a problem in PLS? For a polynomial local search problem, we know that at least one solution (local optimum) must exist. However, many more solutions could exist, how hard is it to count the number of solutions for a PLS-complete problem? I am particularly interested in the decision problem: does this PLS-complete problem's instance have two or more solutions?
Does the complexity depend on which PLS-complete problem we choose? If so then I would be particularly interested in weighted 2SAT (as defined in [SY91] and [Rou10]). I know that counting the number of satisfying solutions for 2SAT is #P-complete, but at first glance it seems like local optima of weighted 2SAT and solutions for 2SAT don't have all that much in common.
I also know that for PLS's cousin PPAD, [CS02] shows that counting the number of Nash equilibria is #P-hard. This suggests that similar PLS problems like counting the number of pure-strategy equilibria in congestion games, would also be hard.
References
[CS02] Conitzer, V., and Sandholm, T. (2002). Complexity results about Nash equilibria. IJCAI-03. cs/0205074.
[Rou10] T. Roughgarden. (2010). Computing equilibria: A computational complexity perspective. Economic Theory, 42:193-236.
[SY91] A.A. Schaeffer and M. Yannakakis. (1991). Simple local search problems that are hard to
solve. SIAM Journal on Computing, 20(1):56-87.
Consider the maximum matching problem in bipartite graphs. The family of feasible solutions consists of all matchings, and local search is performed via finding augmenting paths. The problem belongs to PLS since an augmenting path can be found in polynomial time if a current matching is not maximum, and maximality can be checked in polynomial time. Any local optimum is a maximum matching (i.e., global optimum). However, it is #P-hard to compute the number of maximum matchings in a bipartite graph.

Since a local optimum can be found in polynomial time, the problem is unlikely to be PLS-complete. So, I'm afraid this is not an intended answer (your question restricts to PLS-complete problems). However, I should point out that counting the number of local optima can be hard even though one local optimum can be found efficiently.