$\def\p#1{\left\langle#1\right\rangle}\def\C{\mathbb C}\def\Mat{\operatorname{Mat}}$Let $\p{\cdot,\cdot}$ an inner product on $\C^2$, then there is a positive definite $A \in \Mat_2(\C)$ such that $\p{x,y} = x^tA\bar y$ for all $x,y\in \C^2$. $\p{\cdot,\cdot}$ is $\rho$ invariant, iff for any $x,y$:
$$ x^tA\bar y = \p{x,y} = \p{\rho(z)x,\rho(z)y} = x^t\rho(z)^tA\rho(z)\bar y $$
that is $A = \rho(z)^tA\rho(z)$. We have
\begin{align*}
\rho(z)^tA \rho(z) &= \begin{bmatrix} -1 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}\\
&= \begin{bmatrix} -a-c & -b-d \\ a & b\end{bmatrix} \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}\\
&= \begin{bmatrix} a+b+c+d & -a-c \\ -a-b & a\end{bmatrix}
\end{align*}
That is, we must have
\begin{align*}
a &= a+b+c+d\\
-a-c &= b\\
-a-b &= c\\
a &= d
\end{align*}
which is equivalent to
\begin{align*}
b + c + d &= 0\\
a + b + c &= 0\\
a + b + c &= 0\\
a - d &= 0
\end{align*}
So we must have $a = d$ and $b+c = -a$. As $A$ is hermitian $b = \bar c$, and by definiteness $a > 0$, $ad - bc = a^2 - |b|^2 > 0$. $b+ c = b + \bar b = 2\Re b$. That is $a = -2\Re b$. We need therefore $\Re b < 0$ and
$$ a^2 = 4\Re^2 b > |b|^2 = \Re^2 b + \Im^2 b \iff |\Im b| < -\sqrt 3|\Re b| $$
So the $\rho$-invariant inner products are given by the matrices
$$ \begin{bmatrix} 2\lambda & -\lambda + \mu i \\ -\lambda - \mu i & 2\lambda \end{bmatrix}, \qquad \lambda > 0, |\mu| < \sqrt 3 \lambda $$