Characterization of subgroup of neutral elements of reducible multiary group

Statement
 Consider a reducible multiary group, i.e., a $$n$$-ary group $$G$$ with multiplication $$f:G^n \to G$$ such that there exists a group structure on $$G$$ such that $$f(a_1,a_2,\dots,a_n) = a_1a_2\dots a_n$$ for all (possibly repeated) $$a_1,a_2,\dot,a_n \in G$$, with the multiplication on the right being as per the group structure.

Let $$C$$ be the subset of $$G$$ comprising those elements that are neutral for $$f$$. Consider $$C$$ as a subset of $$G$$ with its (ordinary 2-ary) group operation. The claim is that $$C$$ is a subgroup of $$G$$, and in fact, it is the characteristic subgroup comprising those elements in the center of $$G$$ whose order divides $$n - 1$$.

Note
Note that due to the equivalence of definitions of reducible multiary group, we know that neutral elements exist if and only if the $$n$$-ary group is reducible. Thus, for irreducible $$n$$-ary groups, we would get an empty set instead of a subgroup.

Related facts

 * Equivalence of definitions of reducible multiary group
 * Groups giving same reducible multiary group are isomorphic

Proof
Given: Group $$G$$ with identity element $$e$$. $$C$$ is the subset of $$G$$ comprising those elements $$u$$ of $$G$$ for which $$u^iau^{n-1-i} = a$$ for all $$a \in G$$ and all $$i \in \{ 0,1,\dots, n-1\}$$.

To prove: $$C$$ is precisely the subgroup of $$G$$ comprising the elements $$u$$ in the center of $$G$$ with $$u^{n-1} = e$$.

Proof: