Describing all $\rho$-invariant inner products Let $z$ satisfying the equation $z^3=1$ be a generator of the cyclic group $\mathbb{Z}_3= \{ 1 , z,z^2 \}$. You are given that $\rho : \mathbb{Z}_3 \to GL(\mathbb{C}^2)$ defined by $$\rho(z) = \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}$$ is a representation of the group $\mathbb{Z}_3$ in the vector space $\mathbb{C}^2$.
(a) Find an inner product on the complex vector space $\mathbb{C}^2$ which is $\rho$-invariant: $$\langle \rho(g)u,\rho(g)v \rangle = \langle u,v \rangle$$ for all $g\in\mathbb{Z}_3, u,v\in\mathbb{C}^2$.
(b) Describe all $\rho$-invariant inner products on the vector space $\mathbb{C}^2$ explicitly, in terms of the coordinates of vectors.
I've done (a) but how do I do (b)?
 A: $\def\p#1{\left\langle#1\right\rangle}\def\C{\mathbb C}\def\Mat{\operatorname{Mat}}$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
$$   \begin{bmatrix} 2\lambda & -\lambda + \mu i \\ -\lambda - \mu i  & 2\lambda \end{bmatrix}, \qquad \lambda > 0, |\mu| < \sqrt 3 \lambda $$
A: I think that second $\iff$ in martini's solution should be (no minus sign): $$\text{Im}(b)^2 < 3 \text{Re}(b)^2 \iff |\text{Im}(b)| < \sqrt{3}|\text{Re}(b)|$$ which then gives the matrices as $$\begin{bmatrix} -2\lambda & \lambda + \mu i \\ \lambda - \mu i & -2\lambda \end{bmatrix}$$ for $\lambda <0, |\mu|< \sqrt{3}|\lambda|$ ($\lambda = \text{Re}(b), \mu = \text{Im}(b)$).
These are the same set of matrices.
