Different Coloring of regular n-gon By using Burnside's lemma, I want to find the number of different coloring of vertices of a regular n-gon, with X colors.
By "different" I mean : up to rigid motions.
I've seen some partial results, but I'm not sure about the genralized version.
 A: There are  two possibilities  here, rotational symmetry  (necklace) or
dihedral  symmetry (bracelet).  For the  first one  we have  the cycle
index of the cyclic group:
$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$
For second one we have the cycle index of the dihedral group
$$Z(D_n) = 
\frac{1}{2} Z(C_n) +
\begin{cases} 
\frac{1}{2} a_1 a_2^{(n-1)/2} & n \text{ odd} \\
\frac{1}{4} \left( a_1^2 a_2^{n/2-1} + a_2^{n/2} \right)
& n \text{ even.}
\end{cases}$$
By Burnside  we must  average the  number of  colorings fixed  by each
permutation.  We then use that a permutation fixes a coloring if it is
constant   on  the   cycles,  so   we  have   $X$  choices   for  each
cycle. Therefore we get for necklaces
$$P_n(X) = \frac{1}{n} \sum_{d|n} \varphi(d) X^{n/d}$$
and for bracelets
$$Q_n(X) = 
\frac{1}{2} P_n(X) +
\begin{cases} 
\frac{1}{2} X^{(n+1)/2} & n \text{ odd} \\
\frac{1}{4} \left( X^{n/2+1} + X^{n/2} \right)
& n \text{ even.}
\end{cases}$$
This is for the case of using at most $X$ colors from a set of $X.$ On
the other  hand, if we use  exactly $X$ colors we  have using Stirling
numbers the closed form for necklaces
$$P'_n(X) = \frac{X!}{n} \sum_{d|n} \varphi(d) {n/d\brace X}$$
and for bracelets
$$Q'_n(X) = 
\frac{1}{2} P'_n(X) +
\begin{cases} 
\frac{X!}{2} {(n+1)/2 \brace X} & n \text{ odd} \\
\frac{X!}{4} \left( {n/2+1 \brace X} + {n/2\brace X} \right)
& n \text{ even.}
\end{cases}$$
The Stirling  number formulae  can be derived  by inclusion-exclusion.
This goes as follows.  The nodes $K$  of the poset are all the subsets
of  the colors  $Y$ with  $|Y|=X$ and  represent colorings  using some
subset of the set of colors  $K.$ The weight attached to the colorings
represented at  $K$ is  $(-1)^{|Y|-|K|}.$ Now clearly  colorings using
all colors of $Y$  are included only in the top  node $K=Y$ where they
receive weight  one.  Colorings  using an exact  set $L\subset  Y$ are
represented by all nodes that are supersets of $L$, for a total weight
of
$$\sum_{M\subseteq Y\setminus L} (-1)^{|Y|-(|M|+|L|)}
= \sum_{m=0}^{|Y|-|L|} {|Y|-|L|\choose m} (-1)^{|Y|-(m+|L|)}
\\ = (-1)^{|Y|-|L|}
\sum_{m=0}^{|Y|-|L|} {|Y|-|L|\choose m} (-1)^m = 0.$$
This was zero because $L$ is a  proper subset of $Y.$ We see that when
summing the colorings  represented at all nodes of the  poset only the
ones that use all colors contribute, with a weight of one, so this sum
is the  queried statistic. On  the other  hand summing over  the nodes
first rather than the colorings we obtain
$$\sum_{K\subseteq Y} (-1)^{|Y|-|K|} P_n(|K|)
= \sum_{k=0}^X {X\choose k} (-1)^{X-k} P_n(k)
\\ = \sum_{k=0}^X {X\choose k} (-1)^{X-k}
\frac{1}{n} \sum_{d|n} \varphi(d) k^{n/d}
\\ = \frac{1}{n} \sum_{d|n} \varphi(d) 
\sum_{k=0}^X {X\choose k} (-1)^{X-k} k^{n/d}
\\ = \frac{1}{n} \sum_{d|n} \varphi(d) 
\sum_{k=0}^X {X\choose k} (-1)^{k} (X-k)^{n/d}.$$
We recognize the Stirling number at this point and may conclude. Or if
another step is desired we recall that the combinatorial class for set
partitions is
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}(\mathcal{U} \times \textsc{SET}_{\ge 1}(\mathcal{Z}))$$
giving the EGF
$${n\brace k} = n! [z^n] \frac{(\exp(z)-1)^k}{k!}$$
and note that
$$\sum_{k=0}^X {X\choose k} (-1)^{X-k} k^{n/d}
= (n/d)! [z^{n/d}] 
\sum_{k=0}^X {X\choose k} (-1)^{X-k} \exp(kz)
\\ = (n/d)! [z^{n/d}] (\exp(z)-1)^X
= X! \times (n/d)! [z^{n/d}] 
\frac{(\exp(z)-1)^X}{X!}.$$

Remark. We can  show that the alternate form by  user @Karl is the
same as what we obtain from the cycle index. We get
$$\frac{1}{n} \sum_{k=1}^n X^{\gcd(n,k)}
= \frac{1}{n} \sum_{d|n} \sum_{k=1, \; \gcd(k,n)=d}^n X^d
\\ = \frac{1}{n} \sum_{d|n} X^d \sum_{k=1, \; \gcd(kd,n)=d}^{n/d} 1
= \frac{1}{n} \sum_{d|n} X^d \sum_{k=1, \; \gcd(k,n/d)=1}^{n/d} 1
\\ = \frac{1}{n} \sum_{d|n} \varphi(n/d) X^d.$$
A: Let $C = \{1, ...,X\}^n$ represent the set of colorings of $n$ labeled vertices. The dihedral group $D_n$ acts on $C$ by permuting elements, representing the rigid motions of an $n$-gon. The "distinct colorings" we're looking for are the orbits of this group action.
Burnside's Lemma tells us that the number of orbits is equal to the average number of fixed points of a group element. So for each element $g \in D_n$, we should calculate $|C^g|$, the number of colorings in $C$ that are unchanged by $g$.
$D_n$ consists of $n$ rotations (including the identity element, which we can think of as a rotation by $n$ steps) and $n$ reflections.
If $g$ is a rotation by $k$ steps, then choosing a fixed point of $g$ corresponds to freely choosing the colors of $\gcd(n, k)$ adjacent vertices; the constraint that our coloring is unchanged by $g$ forces us to repeat this color sequence around the polygon, determining the remaining vertices' colors. So $|C^g|=X^{\gcd(n, k)}$.
If $g$ is a reflection, then to build a fixed point we can freely choose the colors of any vertices that lie on the axis of reflection, and the remaining vertices must be colored in pairs so that they match their reflections. If $n$ is odd, each reflection has one vertex on its axis, so $|C^g|=X^{(n+1)/2}$. If $n$ is even, half of the reflections have two vertices on their axis (yielding $|C^g|=X^{n/2+1}$) and the other half have none (yielding $|C^g|=X^{n/2}$).
Putting these things together and taking the average, we find that the number of orbits is
$$
\frac 1{2n}
\left(
\sum_{k=1}^n X^{\gcd(n, k)}
+
\begin{cases}
n X^{(n+1)/2}  & \text{$n$ odd} \\
\frac n 2 X^{n/2+1} + \frac n 2 X^{n/2} & \text{$n$ even}
\end{cases}
\right).
$$
