As a followup to this question (resulting video here), I'd like to make a video showing elements of $\mathbf{SL}(2,\mathbb{R})$ which fix roots of Klein's absolute invariant $j(\tau)$, stylized before and after frames of such a video below:
Given a root $\alpha$ of Klein's absolute invariant $j(\tau)$, which words (in generator matrices $S$ and $T$) of $\mathbf{SL}(2,\mathbb{Z})$ fix $\alpha$?
edit: it's live: Klein's j(τ) whirling upon SL(2,R) with its root exp(2πi/3) fixed
Because of the modular symmetries of $j$, the zeros of $j$ are precisely the $\operatorname{SL}(2,\mathbb{Z})$-transforms of the fundamental-domain zero $\zeta_3=\mathrm{e}^{2\pi\mathrm{i}/3}$.
Let $A\in\operatorname{SL}(2,\mathbb{Z})$ be such that it fixes $\zeta_3$, and let $B\in\operatorname{SL}(2,\mathbb{Z})$ map $\zeta_3$ to another zero $\alpha$ of $j$. Then $M=BAB^{-1}$ fixes $\alpha$. Conversely, if $M\in\operatorname{SL}(2,\mathbb{Z})$ fixes $\alpha$, then $A=B^{-1}MB$ fixes $\zeta_3$. Therefore, the key task is to find all $A\in\operatorname{SL}(2,\mathbb{Z})$ that fix $\zeta_3$.
Note that "$M$ fixes $\alpha$" is equivalent to "$M$ has eigenvector $\begin{pmatrix}\alpha\\1\end{pmatrix}$", but the following reasoning does not need that.
Let $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\operatorname{SL}(2,\mathbb{Z})$, then $$\frac{a\zeta_3+b}{c\zeta_3+d} = \frac{ac+bd-bc + \zeta_3}{c^2-cd+d^2}$$ This shall equal $\zeta_3$. Comparing imaginary and real parts separately, we find that we have to seek integer solutions of $$\begin{aligned} c^2-cd+d^2 &= 1 \\ ac+bd-bc &= 0 \\ ad-bc &= 1 \end{aligned}$$ For the first equation, note that $c=0$ implies $|d|=1$ and vice versa. Furthermore, $c^2-cd+d^2 = (c \pm d)^2 - (1\pm2) cd \geq |cd|$ (choose the $\pm$ such that the right summand becomes nonnegative). These two considerations imply that $c,d$ cannot have absolute value greater than $1$. With the search space thus narrowed, we find $(c,d)\in\{\pm(0,1),\pm(1,0),\pm(1,1)\}$. From that, the remaining two equations can be solved for $a,b$ by writing them as the linear system $$\begin{pmatrix}c&d-c\\d&-c\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}0\\1\end{pmatrix}$$ whose matrix is its own inverse, leading to $$\begin{aligned}a &= d-c & b &= -c\end{aligned}$$ and thus to the solution set $$A\in\left\{ \pm\begin{pmatrix}1&0\\0&1\end{pmatrix}, \pm\begin{pmatrix}-1&-1\\1&0\end{pmatrix}, \pm\begin{pmatrix}0&-1\\1&1\end{pmatrix} \right\} = \left\{\pm I, \pm T^{-1}J, \pm JT\right\}$$ where $$\begin{aligned} I &= \begin{pmatrix}1&0\\0&1\end{pmatrix} & T &= \begin{pmatrix}1&1\\0&1\end{pmatrix} & J &= \begin{pmatrix}0&-1\\1&0\end{pmatrix} \end{aligned}$$ For use as transformations on $\mathbb{H}$, the $\pm$ does not matter. Note that $T^{-1}J = (JT)^2$ and that $(JT)^3=-I$, so the solution set $A$ is the cyclic group generated by $JT$. (You can easily verify that the solution set for any given fixed point $\alpha$ is always a group.) Conjugate the matrices with a suitable $B\in\operatorname{SL}(2,\mathbb{Z})$ to fix zeros of $j$ other than $\zeta_3$.
Edit: Clarified the reasoning a bit.