Does “Second X is NP-complete” imply “X is NP-complete”? "Second $X$" problem is the problem of deciding the existence of another solution different from some given solution for problem instance. 

For some $NP$-complete problems, the second solution version is $NP$-complete (deciding the existence of another solution for the partial Latin square completion problem) while for others it is either trivial (Second NAE SAT) or it can not be $NP$-complete (Second Hamiltonian cycle in cubic graphs) under widely believed complexity conjecture. I am interested in the opposite direction. 

We assume a natural $NP$ problem $X$ where there is natural efficient verifier that verifies a natural interesting relation $(x, c)$ where $x$ is an input instance and $c$ is a short witness of membership of $x$ in $X$. All witnesses are indistinguishable to the verifier. The validity of witnesses must be decided by running the natural verifier and it does not have any knowledge of any correct witness ( both examples in the comments are solutions by definition).  


  Does “Second $X$ is NP-complete” imply “$X$ is NP-complete” for all "natural" problems $X$?


In other words, Are there any "natural" problem $X$ where this implication fails?. Or equivalently,


  Is there any "natural" problem $X$ in $NP$ and not known to be $NP$-complete but its Second $X$ problem is $NP$-complete?


EDIT: Thanks to Marzio's comments, I am not interested in contrived counter-examples. I am only interested in natural and interesting counter-examples for NP-complete problems $X$ similar to the ones above. An acceptable answer is either a proof of the above implication or a counter-example "Second X problem" which is defined for natural, interesting, and well known $NP$ problem $X$.

EDIT 2: Thanks to the fruitful discussion with David Richerby, I have edited the question to emphasis that my interest is only in natural problems $X$.

EDIT 3: Motivation: First, the existence of such implication may simplify the $NP$-completeness proofs of many $NP$ problems. Secondly, the existence of the implication links the complexity of deciding the uniqueness of solution to the problem of deciding existence of a solution for $NP$ problems.
No,

Consider the problem "Find a subset of a set of integers S which sums to 0".

This problem is trivial, as one can return the empty set.

However, finding a second solution after returning the empty set is the well-known subset sum problem, which is known to be NP-complete.