Answer:
I cheated and asked Richard Lyons this question (or at least, the reformulation of the problem, conjecturing that (G,c) is nonsplitting for an involution c with <c> generating G if and only if there exists an odd A such that G/A = Z/2). His response:
Good question! This is a famous (in my circles) theorem - the Glauberman Z^*-Theorem. (Z^*(G) is the preimage of the exponent 2 subgroup of the center of G/O(G), and O(G)=largest normal subgroup of G of odd order.)
Z^*-Theorem: If c is an involution of G then c\in Z^*(G) iff [c,g] has odd order for all g\in G iff for any Sylow 2-subgroup S of G containing c, c is the unique G-conjugate of itself in S.
The last property is absolutely fundamental for CFSG. The proof uses modular character theory for p=2. Attempts to do it with simpler tools have failed.
George Glauberman, Central Elements in Core-free Groups, Journal of Algebra 4, 1966, 403-420.
Older Remarks:
Comment 1: Suppose that P = Z/2+Z/2 is a 2-Sylow. If x lies in P, then P clearly centralizes x, and thus the order of <x> divides #G/P, and is thus odd. By a theorem of Frobenius, G has an odd number of elements of order 2, and thus we see it has an odd number of conjugacy classes of elements of order 2. Yet, by the Sylow theorems, every element of order 2 is conjugate to an element of P. If c lies in P, then by nonsplitting, it is unique in its G-conjugacy class in P. Thus there must be exactly three conjugacy classes of elements of order 2, and thus no element of P is G-conjugate. By a correct application of Frobenius' normal complement theorem, we deduce that G admits a normal subgroup A such that G/A \sim P. Yet <c> generates G, and thus the image of <c> generates G/A. Yet G/A is abelian and non-cyclic, a contradiction.
Comment 2: Suppose that A is a group of order coprime to p such that p | #Aut(A).
Let G be the semidirect product which sits inside the sequence:
1 ---> A ---> G --(phi)--> Z/pZ --> 0;
Let c be (any) element of order p which maps to 1 in Z/pZ. If c is conjugate to
c^j, then phi(c) = phi(c^j). Hence c is not conjugate to any power of itself.
Let H be a subgroup of G containing c (or a conjugate of c, the same argument applies). The element c generates a p-sylow P
of H (and of G). It suffices to show that if gcg^-1 lies in H, then it is conjugate to c inside H. Note that gPg^-1 is a p-Sylow of H.
Since all p-Sylows of H are conjugate, there exists an h such that gPg^-1 = hPh^-1, and thus h c^j h^-1 = gcg^-1. Yet we have seen that c^j is not conjugate to c inside G unless j = 1. Thus gcg^-1 = hch^-1 is conjugate to c inside H.
I just noticed that you wanted <c> to generate G. It's not immediately clear (to me) what condition on A one needs to impose to ensure this. Something like the automorphism has to be "sufficiently mixing". At the very worst, I guess, the group G' generated by <c> still has the property, by the same argument.
This works more generally if p || G and no element of order p is conjugate to a power of itself. (I think you know this already if p = 2.)
The case where the p-Sylow is not cyclic is probably trickier.
Examples: A = (Z/2Z)+(Z/2Z), p = 3. (This is A_4).
A = Quaternion Group, p = 3. (This is GL_2(F_3) = ~A_4, ~ = central extension).
A = M^37, M = monster group, p = 37.
Best Answer
In elementary terms, you have to analyze the following class equation $ n = 1 + h_2 + ... + h_r $ where
Dividing by n, you get $1 = \frac{1}{n} + \frac{1}{c_2} + ... + \frac{1}{c_r} $ which has a finite number of solutions.
Christine Ayoub in her paper On the number of conjugate classes in a group (Proc. Internat. Conf Theory of Groups Canberra 1967) has worked out this analysis for p-groups and there are probably more recent papers on this aspect, which Scott and others allude to in the comments. See for example
Another way of looking at your question is to see that the number of conjugacy classes is the same as the number of irreducible representations. The character table is always square. Therefore, one could ask "what are the number of irreducible characters Irr(G) in a finite group of order n?". The number of linear characters are [G:G'] where G'=commutator subgroup but the nonlinear ones are tougher and there are papers establishing various bounds for these.