<s>[INST] An erroneous application of the Counting Theorem to a regular hexagon? I'm trying to count the unique orbits of a regular hexagon such that each vertex is either Black or White and each edge is either Red, Gree, or Blue. The group I've chosen to act on the hexagon is the dihedral group $D_7$, $$\{e,r,r^2,r^3,r^4,r^5,r^6,s,rs,r^2s,r^3s,r^4s,r^5s,r^6s\}$$ where $r$ is a rotation by $\frac{\pi}{3}$, and $s$ a reflection in the axis connecting two opposite vertices or the midpoints of two opposite edges. When chopped up, I get the following partition into conjugacy classes: $$\{e\} \hspace{0.5cm} \{r,r^6\}\hspace{0.5cm} \{r^2,r^5\}\hspace{0.5cm} \{r^3,r^4\}\hspace{0.5cm}\{s,r^2s,r^4s,r^6s\}\hspace{0.5cm} \{rs,r^3s,r^5s\}$$
Taking the first element of each conjugacy class as the representative, I then go about counting the permutations that are left fixed by that representative. Here's my count (note, $X^g$ denotes the set of all regular hexagons left fixed by group element $g$): $$\begin{align*}
|X^e|&=3^6\times 2^6 & |X^r|&=3\times 2 & |X^{r^2}|&=3^2\times 2^2\\ |X^{r^3}|&=3^3\times 2^3 & |X^s|&=3^4\times 2^3 + 3^3\times 2^4 & |X^{rs}|&=3^3\times 2^4 + 3^4\times 2^3 \end{align*}$$
Notice that the order of the last two sets, $X^s$ and $X^{rs}$, are sums: one addend counts the reflections through opposite vertices and the other through midpoints of opposite sides. When I apply the Counting Theorem (aka Burside's Lemma?) I obtain $$\frac{1}{14}[3^6\times 2^6 + 2(3\times 2) + 2(3^2 \times 2^2) + 2(3^3\times 2^3) + 4(3^4\times 2^3 + 3^3 \times 2^4) + 3(3^3\times 2^4 + 3^4 \times 2^3)]$$ and it is here I stumbled when I saw this product is not an integer.
 [/INST] I now redo the calculation (correctly, I hope), using the notation of the original question. The symmetry group of the hexagon is the dihedral group $D_6$, 
$$ D_6 = \langle r,s \mid r^6 = s^2 = 1, r^s = r^{-1} \rangle, $$ 
where $r$ is a rotation by $\frac{2\pi}{6}$, and $s$ a reflection in an axis connecting two opposite vertices. The conjugacy classes of $D_6$ are 
$$ \{e\}, \hspace{0.5cm} 
   \{r,r^5\}, \hspace{0.5cm} 
   \{r^2,r^4\},\hspace{0.5cm} 
   \{r^3\}, \hspace{0.5cm}
   \{s,r^2s,r^4s\},\hspace{0.5cm} 
   \{rs,r^3s,r^5s\}.$$
For each conjugacy class, we have to count the vertex-and-edge-colored hexagons that are left fixed by a representative of the conjugacy classes.  Let $X^g$ denote the set of all vertex-and-edge-colored regular hexagons left fixed by $g$. Then 
$$\begin{align*}
 |X^e| &= 3^6\times 2^6, & 
 |X^r| &= 3\times 2, & 
 |X^{r^2}| &= 3^2 \times 2^2 ,
 \\ 
 |X^{r^3}| &= 3^3 \times 2^3, & 
 |X^s| &= 3^3 \times 2^4, & 
 |X^{rs}| &= 3^4 \times 2^3 .
\end{align*}$$
For example, $|X^s|= 3^3 \times 2^4$ because $s$ has $3$ orbits on the edges and $4$ orbits on the vertices and so we can choose the colors of $3$ edges and $4$ vertices, such that we get a hexagon that is fixed by $s$. (Recall that $s$ is a reflection in an axis connecting two opposite vertices. $rs$ is a reflection in an axis connecting midpoints of opposite edges.)
Thus the number of orbits is
$$ \frac{1}{12} [3^6 \times 2^6 + 2 (3\times 2) + 2 (3^2 \times 2^2) + (3^3\times 2^3) + 3(3^3 \times 2^4) + 3(3^4 \times 2^3)]  
= 4183.$$
</s>