You should use Burnside's lemma.
There are $4$ rotations of order $12$. Each of these stabilizes $2$ colorings.
There are $2$ rotations of order $6$. Each of these stabilizes $4$ colorings.
There are $2$ rotations of order $4$. Each of these stabilizes $8$ colorings.
There are $2$ rotations of order $3$. Each of these stabilizes $16$ colorings.
There is $1$ rotation of order $2$. Each of these stabilizes $64$ colorings
There is $1$ rotation of order $1$. It stabilizes the $4096$ colorings.
We now apply Burnside and obtain:
$\frac{4\cdot2+2\cdot4+2\cdot8+2\cdot16+1\cdot64+1\cdot 4096}{12}=352$ necklaces.
Paul Raff gave a formula for both bracelets and necklaces so in my answer, I will provide a general method that you can use for this kind of problem. It works also if you want to colour a cube for example.
As Paul Raff pointed out, you did get mix up between bracelet and necklace so in my answer I will include the answer for both of them.
Where did you get it wrong?
It seems to me that you are counting number of ways to colour a bracelet rather than a necklace so I checked your calculation with respect to colouring a bracelet. I would say that the main problem in your counting is that even for each case of the base, you cannot always guarantee to get the final result for that case by multiplying as you did. For example, in the case of $A^2B^2C^2$, we consider two following iterations:
The one on the left gives $\frac{6\cdot5\cdot 4}{3!}=\frac{120}{6}$ ways to choose three colours $A,B,C$, which is what you gave in your table. However, the one on the right gives $\frac{6 \cdot 5 \cdot 4}{2}=\frac{120}{2}$ ways to choose three colours $A,B,C$.
Necklace colouring
You can use Burnside lemma where you can count the number of ways to colour the object by looking at its group of symmetry $G$. For the necklace, the group $G$ can be:
- Two colourings of the necklace are considered the same if from one colouring, we can rotate the necklace to get to the other colouring. There are $5$ possible rotations at angles $60^{\circ}\cdot i \; (i=1,2,3,4,5)$ (not including the do-nothing rotation). Hence, these five rotations are from the mentioned set $G$.
- The "do nothing" action, i.e. we do nothing to the necklace. This is also in $G$.
Let $X$ be the set of all possible colouring for the necklace at a fixed orientation. This follows $|X|=6^6$ as there are $6$ possible colours for each bead.
Now, in Burnside lemma, we essentially want to count number of colourings from $X$ that remains unchanged under actions from $G$. In particular:
- How many colourings in $X$ for the necklace so that it is still the same colouring after we apply $60^{\circ}$ rotation to the necklace? This only happens when all the beads have the same colour. Hence, there are $6$ possible colourings in this case.
The same question can be asked for $120^{\circ}$ rotation: This happens when three of the beads (each is one bead away from the other) have the same colour and the remaining three beads have the same colour.
There are $6$ possible colours for the first three beads and there are $6$ other possible colours for remaining $6$ beads. This gives us $6^2$ possible colourings.
Similarly, with $180^{\circ}$ rotation, a colouring is fixed under this rotation when any pair of opposite beads have the same colour. There are $6$ ways to colour each pair of opposite beads so there are $6^3$ possible colourings.
With $240^{\circ}$ rotation, it's the same as $120^{\circ}$ so we have $6^2$ possible colourings. With $300^{\circ}$, it's the same as $60^{\circ}$ so we have $6$ possible colourings.
With the "do nothing" action then every colouring remains unchanged after this action so there are $6^6$ possible colourings.
The Burnside lemma says that you can add all these numbers up and divide by number of elements of $G$ (which is $6$) to obtain all possible colourings. Hence, the answer for colouring a necklace is
$$\frac{6^6 \cdot 1 +6 \cdot 2+ 6^2 \cdot 2+6^3 \cdot 1}{6}=7826.$$
Bracelet colouring
The difference between bracelets and necklaces is in the group of symmetry $G$. In particular, for bracelets, $G$ has some extra elements: Two colourings of the bracelet are considered same if from one colouring, we can reflect the bracelet through a line to obtain the other colouring. There are two types of lines:
- A line connecting two opposite beads (see diagram on the right). There are $3$ pairs of opposite beads so there are three reflections through these types of lines. These actions are in $G$.
- A line dividing the $6$ beads in equal parts (diagram on the left). There are $3$ such lines corresponding to three extra reflections in $G$.
This time $G$ has $12$ elements.
Next, we do the same thing with necklace, i.e. we count number of colourings that remains fixed under these reflections:
- For reflection across axis on the left, any two beads that appear symmetrically through that axis must have same colour. There are $3$ pairs of such beads so there are $6^3$ possible colourings. Since there are three axes of this type so we have $3 \cdot 6^3$.
- For reflection across axis on the right, note that we can freely choose any colour for the beads that pass through the axis while keeping the same colouring of the bracelet when reflecting. Hence, this gives $6^2$ for such two beads, and $6^2$ for $2$ pairs of symmetrical beads. We obtain $6^4$ possible colourings for such axis. As there are three axes of this type so it is $3 \cdot 6^4$.
Now, applying Burnside's lemma, we sum up all these numbers counted for each element in $G$ then we divide by number of elements in $G$ (which is $12$). The final answer is
$$\frac{1}{12}\left( 6^6 \cdot 1+6 \cdot 2+ 6^2 \cdot 2+6^3 \cdot 1+6^3 \cdot 3+ 6^4 \cdot 3 \right)=4291.$$
Best Answer
We focus our attention on counting length 6 sequences with available entries $\{0,1,2,\dots,9,X\}$.
Simply fix the far left entry of the sequence to be $X$. As there is only ever the one entry with an $X$, this will allow us to treat all rotationally equivalent arrangements the same.
$X~\underline{~}~\underline{~}~\underline{~}~\underline{~}~\underline{~}$
Now... let us ignore the $X$ again and look solely at the five remaining positions. If we were not interested in worrying about reflective symmetry we would find that there are simply $10^5$ possible sequences that can be filled in these remaining positions. As we are concerning ourselves with reflective symmetry however, more care needs to be applied.
We ask ourselves, how many sequences of 5 digits exist unique up to reflection? To count this, we might say that every sequence was counted differently twice: once as $abcde$ and again as $edcba$, so we would have $\frac{10^5}{2}$ such unique sequences however this ignores palindromes.
So, to correct the count, let us first take our $10^5$ sequences, remove the $10^3$ palindromes, divide the result in half to remove duplicates, and then add back in the palindromes.
This gives us a total of $(10^5-10^3)\cdot\frac{1}{2}+10^3=50500$ unique up to reflection sequences of length five. Appending an $X$ at the front of each of these sequences gives us a unique up to reflection and unique up to rotation bracelet.
Side note: Since we were given that every bracelet has exactly one bead $X$, this made things considerably easier for us. We were always able to rotate any bracelet until our $X$ bead was at the far left. Further, when considering reflections, we only ever needed to consider reflecting across the axis containing $X$ since reflection across any other axis would move the $X$ bead from the "left half" to the "right half" of the bracelet or vice versa.
As for what went wrong with your attempt, there were many leaps in logic that I did not understand so I cannot pinpoint everything that you did wrong. One thing that stood out to me was your section:
If we were to generate a necklace ahead of time and try to place our X bead within it, some will have 3 options for where to place X as in your example giving 3 bracelets formed from this necklace, but many others will not.
For example if your original necklace was 00000 then every place where you would insert X would result in the same bracelet giving only one bracelet formed from this necklace.
For another example if your original necklace was 12345 then every place where you would insert X would result in a different bracelet giving six different bracelets formed using this necklace.