Necklace problem with Burnside's lemma 
How many necklaces can be made with two red beads, two green
  beads, and two violet beads?  

I'm trying to solve this with Burnside's Lemma however I'm stuck. Here's what I've got so far:  
Let $S=\{$necklace of length 6 with 2 read, 2 green and 2 violet$\}$, $|S|=\frac{6!}{2!2!2!}=90$, let $G=\mathbb{D}_{6}$, $|G|=2*6=12$.
Now let $g\in G$, $g=e\Rightarrow |fix(g)|=|S|=90$ since $e$ sends any $s$ back to itself.  
For when $g\neq e$, if $g$ is any one of the rotations by $\frac{\pi}{6}$ clockwise, what should I do from here? I intuitively know that not all rotations sends $s$ back to itself and that different configuration of the color beads also plays a role, yet I don't have a clear route to follow.  
And what about flips?  
Thanks for any help!
 A: I upvoted the first answer but I would like to show how to compute the
cycle  index  $Z(D_6)$ of  the  dihedral  group  $D_6$ and  apply  the
Burnside  lemma and  the Polya  Enumeration Theorem  to  this problem.
Observe that  the OEIS uses  the convention of refering  to rotational
symmetry as a necklace and as a bracelet when reflectional symmetry is
included, so we are working with a bracelet here.

We  need  to  enumerate   and  factor  the  twelve  permutations  that
contribute to $Z(D_6).$
There is the identity, which contributes
$$a_1^6.$$
The two rotations by a distance of one and five contribute
$$2 a_6.$$
The two rotations by a distance of two or four contribute
$$2 a_3^2.$$
Finally the rotation by a distance of three contributes
$$a_2^3.$$
There  are three reflections  about an  axis passing  through opposite
vertices, giving
$$3 a_1^2 a_2^2$$
and three reflections  about an axis passing through  the midpoints of
opposite edges, giving
$$3 a_2^3.$$
This finally yields the cycle index
$$Z(D_6) = \frac{1}{12}
\left(a_1^6 + 2a_6 + 2a_3^2 + 3a_1^2 a_2^2 + 4a_2^3\right).$$
Do  the Burnside  calculation  first.  We have  three  colors and  two
instances of each. The colors must  be constant on the cycles.  We now
proceed to count these. We get for $a_1^6$ the contribution ${6\choose
2,2,2}.$  There  are no  candidates  for $a_6$  (we  do  not have  six
instances of a color). We do  not have three instances either, so zero
for $a_3^2.$ For $a_1^2 a_2^2$ we must choose a pair of colors for the
two-cycles, giving $3\times 2\times {3\choose 2}.$ Finally for $a_2^3$
we get six permutations of the three colors. This yields
$$\frac{1}{12}
\left({6\choose 2,2,2} + 3\times 2\times {3\choose 2} 
+ 4\times 6\right)
\\ = \frac{1}{12} (90 + 18 + 24) = 11.$$
On the other hand Polya says that we need the coefficient
$$[A^2 B^2 C^2] Z(D_6)(A+B+C)$$ which is
$$[A^2 B^2 C^2]\frac{1}{12}
\left((A+B+C)^6 + 2(A^6+B^6+C^6) + 
2(A^3 + B^3 + C^3)^2 \\+ 3(A+B+C)^2(A^2+B^2+C^2)^2 
+ 4(A^2+B^2+C^2)^3\right).$$
Now we  may drop the terms  that cannot possibly  produce multiples of
$A^2 B^2 C^2$ which leaves
$$[A^2 B^2 C^2]\frac{1}{12}
\left((A+B+C)^6 
\\ + 3(A+B+C)^2 (A^2+B^2+C^2)^2 
+ 4(A^2+B^2+C^2)^3\right).$$
Doing the coefficient extraction then yields
$$\frac{1}{12}
\left({6\choose 2,2,2}
+ 3 \times 3 \times 2 + 4{3\choose 1,1,1}\right) = 11.$$
Here we  have used the observation  that we must choose  a square from
the first term of $(A+B+C)^2  (A^2+B^2+C^2)^2$ and then choose the two
remaining squares  from the  second factor, which  may be done  in two
ways.
A: Here are the elements of the dihedral group of order twelve:


*

*One identity map.

*Two rotations by a 1/6th turn (clockwise and counterclockwise).

*Two rotations by a 1/3rd turn (clockwise and counterclockwise).

*One rotation by a 1/2 turn.

*Three reflections across lines through two vertices.

*Three reflections across lines through two edges.


For each type of element, look at its "cycle type" as a permutation of the vertices, or in other words the orbit (technically the orbit of the cyclic group it generates).
For instance, let's consider a 1/6th turn. Imagine we have a necklace which is a fixed point of this rotation. The rotation slides the first bead to the second bead position, so they must be the same color, but by the same token it slides the second bead to the third bead position and so on, which implies the beads must all be the same color. This is of course not possible, so $0$ fixed points.
