[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*}[/step]

[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: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*}[/step]

[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: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]