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Are two groups isomorphic iff their cycle index is the same? Note that for every group there exists a permutation group to which it is isomorphic.

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    "Cycle index" is not a function of a group; it's a function of a pair consisting of a group and a permutation representation of it, and it depends on both of these pieces of data.2015-10-18

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No, as per Graphical Enumeration by Hararay and Palmer, p.37:

[two groups] having the same cycle index need not be identical. In fact they need not even be isomorphic. Namely, let p be an odd prime and $m \geq 3$ be an integer ($p=m=3$ is the simplest example). It is well known that there is a nonabelian group of order P^m in which every element except the identity has order p. Let B be the regular representation of this group. Let A be the regular representation of the abelian group of order p^m and type (p,p,...,p). Then A and B are permutation groups of order and degree p^m = d with the same cyclic index $d^{-1}(d_1^d + (d - 1)s^{d/p}_p)$ for each permutation of A and B other than the identity contains p^{m-1} cycles of length p.

This is a quote from a translation of Polya, I can give the full citation if you want (it's in German).

Note that the term "identical," in this context is isomporphism and isomorphism is homomorphism (the reference text is very old).