There may be multiple subgroups that are pairwise permutable complements

Statement
Let $$n$$ be any natural number. Then, there exists a finite group $$G$$ with subgroups $$H_1, H_2, \dots, H_n$$ such that $$H_i$$ and $$H_j$$ are fact about::permutable complements in $$G$$ for any $$i \ne j$$.

Proof
Let $$p$$ be a prime such that $$p + 1 \ge n$$. Let $$G$$ be the group $$\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$$: in other words, $$G$$ is an elementary Abelian group of order $$p^2$$. $$G$$ has $$(p+1)$$ cyclic subgroups of order $$p$$, and any two of these are permutable complements.