elements of $SL(2,\mathbb{Z})$ which fix roots of Klein's absolute invariant $j(\tau)$ 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 
 A: 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.
