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 $\phi(s) = c^{6}, \phi(t) = c.$ Since $G/C$ is Abelian, ${\rm ker \phi}$ contains the derived group $[G,G]$ in its kernel so all $G$-conjugates of $t$ have the same image $c$ under $\phi$.

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 $\phi$.

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 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 $\phi(s) = c^{6}, \phi(t) = c.$ Since $G/C$ is Abelian, ${\rm ker \phi}$ contains the derived group $[G,G]$ in its kernel so all $G$-conjugates of $t$ have the same image $c$ under $\phi$.

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 $\phi(s) = c^{6}, \phi(t) = c.$ Since $G/C$ is Abelian, ${\rm ker \phi}$ contains the derived group $[G,G]$ in its kernel so all $G$-conjugates of $t$ have the same image $c$ under $\phi$.