Let the language $L$ consist of the $k$-CNF formulas $\phi$ with the property that any satisfying assignment $x$ of $\phi$ is a Not-All-Equal (NAE) assignment, i.e. every clause of $\phi$ has at least one false literal under $x$. Equivalently, $L$ includes those $k$-CNF formulas $\phi$ for which $\phi(x) \implies \phi(\neg x)$.
Clearly $L$ is in coNP, since a certificate for a "no" instance is given by a satisfying assignment which is not a NAE assignment. Is $L$ also coNP-complete for $k > 2$?
As a simple example consider the $2$-CNF formula $(a \vee b) \wedge (\neg a \vee \neg b)$ whose satisfying solutions have exactly one of $a$ or $b$ true, but not both.