$C_{12}$ is the cyclic group of order 12. So it's not $D_{12}$ (aka $D_{24}$) which has order 24.
So, looking at the two elements you've identified:
- $e$, or "do nothing". That doesn't have order 12.
- "rotate $30^\circ$", which you've written as $(12,1,2,3,4,5,6,7,8,9,10,11)$.
Let's call that one $a$. "rotate by $30^\circ$" has order 12, and the group it generates is therefore $C_{12}$, the cyclic group of order $12$.
You could identify all the elements of $\langle a\rangle$ which have order $12$. The elements themselves are easy to identify, they are $e=a^0$, $a=a^1$, $a^2$, $a^3$, $a^4$, etc up to $a^{11}$ (since $a^{12}=e=a^0$).
Note that I also want to point out that when you wrote $e$ and $a$ above, you didn't write them in "cycle notation". Cycle notation might look like this:
- $p=(1,4,6,2)(3,7)$ which means $p$ maps $1\rightarrow4$, $2\rightarrow1$, $3\rightarrow7$, $4\rightarrow6$, $5\rightarrow5$, $6\rightarrow2$, $7\rightarrow3$, and for $n\geq 8$, $n\rightarrow n$.
So $e$, in cycle notation, is just $()$. Your $a$, which I think you have mapping $1$ to $12$, would actually be written $(1,12,11,10,9,8,7,6,5,4,3,2)$, or (equivalently), $(6,5,4,3,2,1,12,11,10,9,8,7)$ or a few other possibilities, depending what you choose to "start" the cycle with.
Then, $a^2$ would map $1\rightarrow11$, $11\rightarrow9$, $9\rightarrow7$, etc, so you'd write it as $(1,11,9,7,5,3)(2,12,10,8,6,4)$. That has order $6$. I can tell, because $6$ is the lowest common multiple of the lengths of all the cycles in $a^2$.
Write $a$ in cycle notation, then work out $a^2$, $a^3$, $a^4$ etc up to $a^{11}$, and identify which ones have order $12$. There should be four of them.
