TITLE: How many $n$-pointed stars are there?
QUESTION [7 upvotes]: Say we have $n$ distinct points spaced evenly in a circle. Define a star as a connected graph with these points as vertices and with $n$ edges, no two having the same endpoints. We think of two stars as being equivalent if they differ only by a rotation (if they differ by a reflection I consider them as different objects). I'd like to know how many different stars there are on $n$ points. 
For example there are only two 4-pointed stars: a square and a bowtie. There are four 5-pointed stars:

And I think there are twelve 6-pointed stars:

(I'm not completely sure that I have exhausted the possibilities for 6 points).
A quick search of the online encyclopedia of integer sequences hasn't revealed any obvious candidates.
Another way to phrase the question:
Let $\rho = (1 2 \ldots n)\in S_n$. Say that two permutations $\sigma, \tau\in S_n$ are equivalent if $\sigma = \rho^{-k}\tau\rho^k$ for some $k$. How many equivalence classes contain order $n$ permutations?

REPLY [4 votes]: Deriving the general formula for this sequence (and its reflection-insensitive analogue) is precisely the content of:

Golomb, S.W.; Welch, L.R., "On the enumeration of polygons", Amer. Math. Monthly, 67 (1960), 349-353.

The authors derive the result as a clever application of Burnside's Lemma, giving for the count of $n$-pointed stars
$$
E(n)
=
\left\{
\begin{array}{ll}
F(n) , & \textrm{$n$ odd}\\
F(n) + \displaystyle{\frac{1}{2 n^2} \cdot 2^{n / 2} {n \choose 2} {n \choose 2}!}, & \textrm{$n$ even}
\end{array}
\right. ,
$$
where
$$F(n) := \frac{1}{2 n^2} \sum_{d \mid n} \phi^2\left(\frac{n}{d}\right) d! \left(\frac{n}{d}\right)^d .$$
Here, $\phi$ is the Euler totient function.
As Dan Uznanski's answer has pointed out, this is OEIS A000939, Number of inequivalent n-gons.
Remark As Empy2's answer observes, the procedure for computing should be easier for prime $n$. Indeed, for odd primes $p$ the above formula simplifies to
$$E(p) = \frac{1}{2 p} [(p - 1)^2 + (p - 1)!] .$$ It follows from the fact that $E(p)$ is an integer that $p \mid [(p - 1)^2 + (p - 1)!]$, and rearranging gives $$(p - 1)! \equiv -1 \pmod p ,$$ which recovers one direction of Wilson's Theorem.