Groups giving same reducible multiary group are isomorphic

Statement
 Suppose $$G$$ is a $$n$$-ary group (i.e., a multiary group with arity $$n$$) with $$n$$-ary operation $$f$$ that is reducible. Suppose $$*$$ and $$\cdot$$ are multiplications on $$G$$, both making $$G$$ into a group, that both induce the $$n$$-ary operation $$f$$ when thought of the usual way:

$$f(a_1,a_2,\dots,a_n) = a_1 * a_2 * \dots * a_n = a_1 \cdot a_2 \cdot \dots \cdot a_n$$ for all $$a_1,a_2,\dots,a_n \in G$$

Then, the groups $$(G,*)$$ and $$(G,\cdot)$$ are isomorphic groups, and the isomorphism can be expressed explicitly in either group in terms of multiplication by a central element of order dividing $$n - 1 $$.

Related facts

 * Equivalence of definitions of reducible multiary group
 * Characterization of subgroup of neutral elements of reducible multiary group

Facts used

 * 1) uses::Characterization of subgroup of neutral elements of reducible multiary group

Proof
Given: Set $$G$$ with two group operations $$*$$ and $$\cdot$$ such that:

$$a_1 * a_2 * \dots * a_n = a_1 \cdot a_2 \cdot \dots \cdot a_n$$ for all $$a_1,a_2,\dots,a_n \in G$$

To prove: $$(G,*)$$ and $$(G,\cdot)$$ are isomorphic groups

Proof: For simplicity, we will denote $$\cdot$$ by concatenation but write $$*$$ explicitly. Let $$e$$ be the identity element for $$\cdot$$ and $$u$$ be the identity element for $$*$$. $$u$$ is a neutral element for $$f$$, hence by Fact (1), it lies in the center of $$G$$ and $$u^{n-1} = e$$. Further, by the equality of operations:

$$a * b = a * u * \dots * u * b = au^{n-2}b$$

Using that $$u^{n-1} = e$$ and that $$u$$ is in the center, this gives:

$$a * b = aub$$

Multiplying both sides by $$u$$ (this is with respect to $$\cdot$$, the default multiplication):

$$u(a * b) = (ua)(ub)$$

We are now in a position to define the isomorphism $$(G,*) \to (G,\cdot)$$. The isomorphism is:

$$x \mapsto u \cdot x$$

The above shows that it is a homomorphism. It is clearly bijective, hence it is an isomorphism.