Let $G \subseteq GL_d (\mathbb C)$ be a finite pseudoreflection group (see here and here) acting on a domain $\Omega \subseteq \mathbb C^d$ by the right action $\sigma \cdot z = \sigma^{-1} z$ where $\sigma \in G$ and $z \in \mathbb \Omega.$ This gives rise to a left group action of $G$ on the space of all $\mathbb C^n$-valued functions defined on $\Omega$ given by $\sigma (f) (z) = f(\sigma^{-1} z).$ A function $f : \Omega \longrightarrow \mathbb C^n$ is said to be $G$-invariant if $\sigma (f) = f$ for all $\sigma \in G$; while a function $f : \Omega \longrightarrow \mathbb C^n$ is said to be precisely $G$-invariant if it is $G$-invariant and $f(x) = f(y)$ if and only if $x$ and $y$ are in the same $G$-orbit i.e. there exists some $\sigma \in G$ such that $y = \sigma^{-1} x.$ Now by Chevalley-Shephard-Todd Theorem we know that the ring of $G$-invariant polynomials is again a polynomial ring i.e. there exist algebraically independent homogeneous $G$-invariant polynomials $\theta_1, \cdots, \theta_d$ (not necessarily unique) such that $\mathbb C[z]^G = \mathbb C[\theta_1, \cdots, \theta_d].$ Then $\theta = (\theta_1, \cdots, \theta_d) : \mathbb C^d \longrightarrow \mathbb C^d$ is known as a basic polynomial map associated to the pseudoreflection group $G.$ Rudin showed in his paper Proper Holomorphic Maps and Finite Reflection Groups (Proposition 2.2) that such a map is proper and precisely $G$-invariant. Now I would like to find out the group of deck transformations of the basic polynomial map. In Proposition 5.4 in this (or this) paper it is claimed that the group of deck transformations of the basic polynomial map $\theta : \Omega \longrightarrow \theta (\Omega)$ is $G$ itself.
I believe that it is meant to say that the group of deck transformations of $\theta$ is isomorphic to $G.$ Since $\theta$ is $G$-invariant it is clear that the biholomorphisms given by $z \mapsto \sigma^{-1} z$ ($z \in \Omega$) is a deck transformation of $\theta.$ On the other hand if $f$ is a deck transformation of $\theta$ then $\theta (f(z)) = \theta (z)$ for all $z \in \Omega.$ Hence by precise $G$-invariance property of $\theta$ it follows that $f(z)$ and $z$ should lie in the same $G$-orbit. So there exists $\sigma_z \in G$ (depending upon $z \in \Omega$) such that $f(z) = \sigma_z^{-1} z,\ z \in \Omega.$ Can the choice of $\sigma_z \in G$ be made independent of the choice of $z \in \Omega$ so that the deck transformations of $\theta$ will precisely be of the form
$$f_{\sigma} : \Omega \ni z \mapsto \sigma^{-1} z \in \Omega.$$ Any suggestion in this regard would be warmly appreciated. Thanks for your time.
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