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.

O negligence! Fit for a fool to fall by.
All credit to @ladisch for obviously making the obvious.
a) I should have used $D_6$ instead of $D_7$. So chop off the $r^6$ and $r^6s$ from the group and we get the dihedral group of order 12.
b) The conjugacy classes containing rotations only are adjusted acrodingly and $\{s,r^2s,r^4s,r^6s\}$ becomes $\{s,r^2s,r^4s\}$.
c) My counting of the elements of $X^g$, where $g$ is a representative of the conjugacy classes, remains the same...
d)... but the product $$\frac{1}{12}[3^6\times 2^6 + 2(3\times 2) + 3^2 \times 2^2 + 2(3^3\times 2^3) + 3(3^4\times 2^3) + 3(3^3\times 2^4)]=4183$$
