Example 7.10 Let $A_{6}$ be a regular hexagon. Now we use red and blue to color the 6 vertices of $A_{6}$, with each vertex being one color. Find the number of type II patterns of the vertex-colored regular hexagon.
Solution: Let $(D, *)$ be the non-normal rotation group of $A_{6}$, and the vertex permutation group it induces is $(F, \circ)$. Suppose $d \in D$, and the vertex permutation induced by $d$ is $f$. Let $A_{6}^{\prime}$ be the vertex-colored hexagon obtained by coloring the 6 vertices of $A_{6}$ with two colors, red and blue. Then $d A_{6}^{\prime} = A_{6}^{\prime}$ if and only if the vertices of $A_{6}^{\prime}$ in the same cycle of $g$ have the same color. From Example 7.5 of this chapter, we have
$$
P_{F}\left(x_{1}, x_{2}, \cdots, x_{6}\right)=\frac{1}{12}\left(x_{1}^{6}+4 x_{2}^{3}+2 x_{3}^{2}+3 x_{1}^{2} x_{2}^{2}+2 x_{6}\right),
$$

Therefore, by Burnside's Lemma, the number of type II vertex-colored hexagons is
$$
N=\frac{1}{12}\left(2^{6}+4 \times 2^{3}+2 \times 2^{2}+3 \times 2^{2} \times 2^{2}+2 \times 2\right)=13 .
$$