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.
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:
Then to generate unique bracelets from these necklaces, I have three options to place the X bead for each. Ex. Original Necklace = 00001
Unique Bracelets Generated = X00001, 0X0001, 00X001
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.
Best Answer
Let us consider this problem from a graph theory perspective.
Consider the bracelets made up of five beads, which can be of red, blue or green colour. Then let us have $D =\{1,2,3,4,5\}$ is the set of places the beads can take, $C =\{r, b, g\}$ is the set of colours that a bead in a particular place can take and $C_5$ is the cyclic group of order $5$ generated by the permutation $(12345)$. Then the question is: how many orbits does $C_5$ have?. For the identity, $1\in C_5$, clearly, $F(1) = X$.
For a permutation $\alpha \in C_5$, we denote by $F(\alpha)$ the set of elements fixed by $\alpha$ as $F(\alpha) = \{x\in X: \alpha x=x\}$ where $X$ is considered to be the set of all possible bracelets (which are $3^5 = 243$ in number).
Now, we use the Cauchy-Frobenius lemma to find the number of orbits. The original form of the lemma says that, $$N(\tau) =\frac{1}{|\tau|} \sum_{ \alpha \in \tau} |F (\alpha)|$$ where $N(\tau)$ is the number of orbits.
Returning to our problem, we know that for every non-trivial rotation $\alpha \in C_5$, only the three monochromatic bracelets are invariant under $\alpha$, so $$N(\tau) = \frac{1}{5}[243 + 3 + 3 + 3 + 3] = 51$$ Hope it helps.
Reference: Modern Graph Theory by Bella Bollobas (pg.$277-279$)