Necklaces over a group with identity product

We address two variants of the classical necklace counting problem from enumerative combinatorics. In both cases, we fix a finite group G and a positive integer n. In the first variant, we count the ``identity-product n-necklaces'' -- that is, the orbits of n-tuples (a₁, a₂, …, aₙ) ∈ Gⁿ that satisfy a₁ a₂ ⋯ aₙ = 1 under cyclic rotation. In the second, we count the orbits of all n-tuples (a₁, a₂, …, aₙ) ∈ Gⁿ under cyclic rotation and left multiplication (i.e., the operation of G on Gⁿ given by h · (a₁, a₂, …, aₙ) = (ha₁, ha₂, …, haₙ)). We prove bijectively that both answers are the same, and express them as a sum over divisors of n.