Is there a modal operator, L, that satisfies φ ↔ & ~L ⊢ ~Lφ?

Question

I am wondering if there's some modal operator that would satisfy $$φ ↔ \& ~L ⊢ ~Lφ.$$ That is:

1. Given $φ ↔ \& ~L$
2. You can get to $~Lφ$

One limitation is that $L$ for sure does not satisfy modal axiom $T$: i.e., it's not the case that if $Lφ → φ$.

Otherwise, we'd use the usual, well-known modal axioms and theorems.

Thanks for all your help.