Does "$\forall x\in L, \sigma(\neg x)=\neg \sigma(x)$" hold given that $\sigma(F)\equiv F$ for a CNF formula $F$ built on a set $L$ of literals?
No.  Suppose $F$ is the formula

$$(x_1 \lor x_2) \land (\neg x_1 \lor x_2) \land (x_1 \lor \neg x_2) \land (\neg x_1 \lor \neg x_2) \land (x_1 \lor \neg x_1) \land (x_2 \lor \neg x_2),$$

and consider the permutation $\sigma$ such that $\sigma(x_1)=x_2$, $\sigma(x_2)=\neg x_1$, $\sigma(\neg x_1)=x_1$, and $\sigma(\neg x_2)=\neg x_2$.

Then $\sigma(F) \equiv F$, but $\sigma(\neg x_1) \ne \neg \sigma(x_1)$.