[proofplan]
We compute the stabilizer equations directly from the fractional linear action. For $i$, the equation $A \cdot i=i$ forces a matrix of the form $\begin{pmatrix}a&-c\\ c&a\end{pmatrix}$, and the determinant condition reduces to $a^2+c^2=1$. For $\rho$, the identity $\rho^2+\rho+1=0$ lets us compare coefficients in the $\mathbb{Q}$-basis $\{1,\rho\}$ of the field $\mathbb{Q}(\rho):=\{u+v\rho:u,v\in\mathbb{Q}\}$, where $\mathbb{Q}$ denotes the field of rational numbers, reducing the problem to the integer equation $a^2+ac+c^2=1$. Finally, passing to $PSL_2(\mathbb{Z})$ identifies each matrix with its negative.
[/proofplan]
[step:Compute the stabilizer equation at $i$]
Let
\begin{align*}
A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb{Z}),
\end{align*}
so $a,b,c,d\in \mathbb{Z}$ and $ad-bc=1$. The condition $A\cdot i=i$ is
\begin{align*}
\frac{ai+b}{ci+d}=i.
\end{align*}
Since $ci+d\neq 0$, multiplying by $ci+d$ gives
\begin{align*}
ai+b=i(ci+d)=ci^2+di=-c+di.
\end{align*}
Comparing real and imaginary parts gives
\begin{align*}
b=-c, \qquad a=d.
\end{align*}
Thus every element of the stabilizer of $i$ has the form
\begin{align*}
A=\begin{pmatrix}a&-c\\c&a\end{pmatrix}.
\end{align*}
The determinant condition becomes
\begin{align*}
1=\det A=a^2+c^2.
\end{align*}
Since $a,c\in \mathbb{Z}$, the integer solutions are
\begin{align*}
(a,c)\in \{(1,0),(-1,0),(0,1),(0,-1)\}.
\end{align*}
These give precisely
\begin{align*}
I,\quad -I,\quad S,\quad -S.
\end{align*}
Therefore
\begin{align*}
(SL_2(\mathbb{Z}))_i=\{\pm I,\pm S\}.
\end{align*}
[guided]
Let
\begin{align*}
A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb{Z}).
\end{align*}
By definition, this means $a,b,c,d\in\mathbb{Z}$ and
\begin{align*}
ad-bc=1.
\end{align*}
We want to determine exactly when $A$ fixes $i$ under the fractional linear action. The condition is
\begin{align*}
A\cdot i=i,
\end{align*}
which means
\begin{align*}
\frac{ai+b}{ci+d}=i.
\end{align*}
The denominator $ci+d$ is nonzero because otherwise $d=-ci$, forcing $c=d=0$ since $c,d\in\mathbb{Z}$, contradicting $ad-bc=1$. Hence we may multiply by $ci+d$:
\begin{align*}
ai+b=i(ci+d).
\end{align*}
Now compute the right-hand side using $i^2=-1$:
\begin{align*}
i(ci+d)=ci^2+di=-c+di.
\end{align*}
Thus
\begin{align*}
ai+b=-c+di.
\end{align*}
Equality of complex numbers forces equality of real and imaginary parts, so
\begin{align*}
b=-c,\qquad a=d.
\end{align*}
Therefore any stabilizing matrix has the form
\begin{align*}
A=\begin{pmatrix}a&-c\\c&a\end{pmatrix}.
\end{align*}
Now we use the condition $A\in SL_2(\mathbb{Z})$. Its determinant is
\begin{align*}
\det A=a\cdot a-(-c)c=a^2+c^2.
\end{align*}
Since $\det A=1$, we get
\begin{align*}
a^2+c^2=1.
\end{align*}
The only integer pairs satisfying this equation are
\begin{align*}
(a,c)\in \{(1,0),(-1,0),(0,1),(0,-1)\}.
\end{align*}
Substituting these four pairs gives
\begin{align*}
\begin{pmatrix}1&0\\0&1\end{pmatrix}=I,\qquad
\begin{pmatrix}-1&0\\0&-1\end{pmatrix}=-I,
\end{align*}
and
\begin{align*}
\begin{pmatrix}0&-1\\1&0\end{pmatrix}=S,\qquad
\begin{pmatrix}0&1\\-1&0\end{pmatrix}=-S.
\end{align*}
Thus the stabilizer of $i$ in $SL_2(\mathbb{Z})$ is exactly
\begin{align*}
(SL_2(\mathbb{Z}))_i=\{\pm I,\pm S\}.
\end{align*}
[/guided]
[/step]
[step:Compute the stabilizer equation at $\rho$]
Let
\begin{align*}
A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb{Z}).
\end{align*}
The point $\rho$ satisfies
\begin{align*}
\rho^2+\rho+1=0.
\end{align*}
The condition $A\cdot \rho=\rho$ is
\begin{align*}
\frac{a\rho+b}{c\rho+d}=\rho.
\end{align*}
The denominator $c\rho+d$ is nonzero: if $c\rho+d=0$ and $c\neq 0$, then $\rho=-d/c\in\mathbb{Q}$, while if $c=0$, then $d=0$, contradicting $ad-bc=1$. Therefore multiplication by $c\rho+d$ gives
\begin{align*}
a\rho+b=c\rho^2+d\rho.
\end{align*}
Using $\rho^2=-\rho-1$, the right-hand side becomes
\begin{align*}
c\rho^2+d\rho=-c+(d-c)\rho.
\end{align*}
Here $\mathbb{Q}$ denotes the field of rational numbers, and $\mathbb{Q}(\rho):=\{u+v\rho:u,v\in\mathbb{Q}\}$ denotes the field generated by $\rho$ over $\mathbb{Q}$. Since $\rho\notin\mathbb{Q}$, the elements $1$ and $\rho$ are linearly independent over $\mathbb{Q}$. Comparing coefficients in the $\mathbb{Q}$-basis $\{1,\rho\}$ gives
\begin{align*}
b=-c,\qquad a=d-c.
\end{align*}
Equivalently,
\begin{align*}
d=a+c,\qquad b=-c.
\end{align*}
Hence every element of the stabilizer of $\rho$ has the form
\begin{align*}
A=\begin{pmatrix}a&-c\\c&a+c\end{pmatrix}.
\end{align*}
The determinant condition is
\begin{align*}
1=\det A=a(a+c)-(-c)c=a^2+ac+c^2.
\end{align*}
The integer solutions of
\begin{align*}
a^2+ac+c^2=1
\end{align*}
are
\begin{align*}
(a,c)\in \{(1,0),(-1,0),(0,1),(0,-1),(1,-1),(-1,1)\}.
\end{align*}
Substituting these six pairs gives
\begin{align*}
I,\quad -I,\quad ST,\quad -ST,\quad -(ST)^2,\quad (ST)^2.
\end{align*}
Therefore
\begin{align*}
(SL_2(\mathbb{Z}))_\rho=\{\pm I,\pm ST,\pm (ST)^2\}.
\end{align*}
[guided]
Let
\begin{align*}
A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb{Z}).
\end{align*}
We again translate the fixed-point condition into equations for the integer entries of $A$. The point
\begin{align*}
\rho=\frac{-1+i\sqrt{3}}{2}
\end{align*}
is a root of
\begin{align*}
x^2+x+1=0,
\end{align*}
so
\begin{align*}
\rho^2+\rho+1=0.
\end{align*}
The condition that $A$ fixes $\rho$ is
\begin{align*}
\frac{a\rho+b}{c\rho+d}=\rho.
\end{align*}
Let $\mathbb{Q}$ denote the field of rational numbers, and let $\mathbb{Q}(\rho):=\{u+v\rho:u,v\in\mathbb{Q}\}$ denote the field generated by $\rho$ over $\mathbb{Q}$. The denominator $c\rho+d$ is nonzero. Indeed, if $c\rho+d=0$, then $\rho=-d/c$ when $c\neq 0$, contradicting $\rho\notin\mathbb{Q}$; if $c=0$, then $d=0$, contradicting $ad-bc=1$. Hence multiplication by $c\rho+d$ is valid, and we obtain
\begin{align*}
a\rho+b=\rho(c\rho+d)=c\rho^2+d\rho.
\end{align*}
Now use the quadratic relation $\rho^2=-\rho-1$:
\begin{align*}
c\rho^2+d\rho=c(-\rho-1)+d\rho=-c+(d-c)\rho.
\end{align*}
Thus
\begin{align*}
b+a\rho=-c+(d-c)\rho.
\end{align*}
The two elements $1$ and $\rho$ are linearly independent over $\mathbb{Q}$ because $\rho$ is not rational. Therefore equality in the $\mathbb{Q}$-basis $\{1,\rho\}$ of $\mathbb{Q}(\rho)$ forces equality of the constant coefficients and of the $\rho$-coefficients:
\begin{align*}
b=-c,\qquad a=d-c.
\end{align*}
Equivalently,
\begin{align*}
b=-c,\qquad d=a+c.
\end{align*}
So any stabilizing matrix must have the form
\begin{align*}
A=\begin{pmatrix}a&-c\\c&a+c\end{pmatrix}.
\end{align*}
We now impose the condition $A\in SL_2(\mathbb{Z})$. Its determinant is
\begin{align*}
\det A=a(a+c)-(-c)c=a^2+ac+c^2.
\end{align*}
Thus
\begin{align*}
a^2+ac+c^2=1.
\end{align*}
To solve this over $\mathbb{Z}$, rewrite
\begin{align*}
4(a^2+ac+c^2)=(2a+c)^2+3c^2.
\end{align*}
Since $a^2+ac+c^2=1$, this gives
\begin{align*}
(2a+c)^2+3c^2=4.
\end{align*}
Hence $3c^2\leq 4$, so $c\in\{-1,0,1\}$. Checking these three cases:
\begin{align*}
c=0 &\implies a^2=1 \implies a=\pm 1,\\
c=1 &\implies a^2+a+1=1 \implies a(a+1)=0 \implies a\in\{0,-1\},\\
c=-1 &\implies a^2-a+1=1 \implies a(a-1)=0 \implies a\in\{0,1\}.
\end{align*}
Thus the integer solutions are
\begin{align*}
(a,c)\in \{(1,0),(-1,0),(0,1),(0,-1),(1,-1),(-1,1)\}.
\end{align*}
Substituting these into
\begin{align*}
A=\begin{pmatrix}a&-c\\c&a+c\end{pmatrix}
\end{align*}
gives
\begin{align*}
(1,0)&\mapsto I,\\
(-1,0)&\mapsto -I,\\
(0,1)&\mapsto \begin{pmatrix}0&-1\\1&1\end{pmatrix}=ST,\\
(0,-1)&\mapsto \begin{pmatrix}0&1\\-1&-1\end{pmatrix}=-ST,\\
(1,-1)&\mapsto \begin{pmatrix}1&1\\-1&0\end{pmatrix}=-(ST)^2,\\
(-1,1)&\mapsto \begin{pmatrix}-1&-1\\1&0\end{pmatrix}=(ST)^2.
\end{align*}
Therefore
\begin{align*}
(SL_2(\mathbb{Z}))_\rho=\{\pm I,\pm ST,\pm (ST)^2\}.
\end{align*}
[/guided]
[/step]
[step:Pass from $SL_2(\mathbb{Z})$ to $PSL_2(\mathbb{Z})$]
The group $PSL_2(\mathbb{Z})$ is the quotient
\begin{align*}
PSL_2(\mathbb{Z})=SL_2(\mathbb{Z})/\{\pm I\}.
\end{align*}
Thus each pair $\{A,-A\}$ in $SL_2(\mathbb{Z})$ determines a single element of $PSL_2(\mathbb{Z})$.
For $i$, the stabilizer in $SL_2(\mathbb{Z})$ is
\begin{align*}
\{\pm I,\pm S\},
\end{align*}
which has two pairs modulo signs:
\begin{align*}
\{\pm I\},\qquad \{\pm S\}.
\end{align*}
Hence the stabilizer of $i$ in $PSL_2(\mathbb{Z})$ has order $2$.
For $\rho$, the stabilizer in $SL_2(\mathbb{Z})$ is
\begin{align*}
\{\pm I,\pm ST,\pm (ST)^2\},
\end{align*}
which has three pairs modulo signs:
\begin{align*}
\{\pm I\},\qquad \{\pm ST\},\qquad \{\pm (ST)^2\}.
\end{align*}
Hence the stabilizer of $\rho$ in $PSL_2(\mathbb{Z})$ has order $3$. This proves all asserted stabilizer descriptions.
[/step]
