Product of conjugate matrices in $\mathrm{SL}(2, \mathbb{Z})$

See Keith Conrad's notes http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/SL(2,Z).pdf, particularly Example 2.5. Let us write (as Conrad does) $S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and $T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. Then $S$ has order $4$ and $ST = \begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix}$ has order $6$. In his Example 2.5, Conrad displays a character $\chi : SL_2(\mathbb{Z}) \to \mathbb{C}^*$ taking $S$ to a primitive $4$th root of unity and $ST$ to a primitive $6$th root of unity. So $\chi(T) = \chi(S)^{-1} \chi(ST)$ is a primitive $12$th root of unity; in fact Conrad's example has $\chi(T) = e^{2\pi i/12}$. And $\chi(M)$ has this same value for any matrix $M$ conjugate to $T$.

At this point perhaps it is obvious, but to be clear: if $M_1,\dotsc,M_n$ are each conjugate to $T$ and $M_1 \dotsm M_n$ is the identity, then $1 = \chi(\prod M_i) = \prod \chi(M_i) = \chi(T)^n = e^{2 \pi i \, n/12}$, so $12 \mid n$.

To expand my comment, and combine it with some points from Zach Teitler's answer, but more in a generators and relations framework: ${\rm SL}(2,\mathbb{Z})$ is well-known to be isomorphic to the group $G = \langle s,t: s^{4} = (st)^{6} = 1, s^{2} = (st)^{3} \rangle $. But notice that the cyclic group $C = \langle c \rangle $ of order $12$ satisfies these relations ( with an element of order $4$ in the role of $s$ and an element of order $12$ in the role of $t$).

Hence there is a homomorphism $\phi: G \to C$ with $\phi(s) = c^{-3}, \phi(t) = c.$ Since $G/C$ is Abelian, $\phi$ contains the derived group $[G,G]$ in its kernel so all $G$-conjugates of $t$ have the same image $c$ under $\phi$.