I have a question about the following remark 5.6 of this article, I'm reading right now. I try to give a little pretext, with the notations used in the paper:
Definition 5.2: Let $H(a; b) \in GL(n, \mathbb{C})$ be a hypergeometric group with parameters $a_1, ..., a_n, ; b_1 ..... b_n$, and generators $h_0, h_1, h_{\infty}$ as in Definition 3.1. The subgroup $H_r(a;b)$ of $H(a;b)$ generated by the reflections $h_{\infty}^k h_1 h_{\infty}^{-k}$ for $k\in \mathbb{Z}$ is called reflection subgroup of $H(a; b)$.
Definition 5.5: A scalar shift of the hypergeometric group $H (a; b)$ is a hypergeometric group $H (da; db)= H (da_1, ..., da_n ; db_1, ..., db_n)$ for some $d\in \mathbb{C}^*$.
Remark 5.6: If $d$ has the form $d=exp(2\pi i \delta)$ for some $\delta \in \mathbb{C}$ then a scalar shift from $H(a; b)$ to $H(da; db)$ is the effect on the monodromy group obtained by multiplying all solutions of the hypergeometric equation by $z^{-\delta}$. Observe that the associated reflection groups $H_r(a; b)$ and $H_r(da; db)$ are naturally isomorphic.
My question and what i have done so far:
As said before, my question is about remark 5.6, or to be more precise, about the "second" part, regarding the natural isomorphism. I still want to at least explain my progress on the first part of 5.6 so far:
First note, that the monodromy group of a hypergeometric equation $D(\alpha;\beta)u=0$ is a hypergeometric group $H(a;b)$ with parameters $a_j=exp(2 \pi i \alpha_j)$ and $b_j = exp(2 \pi i \beta_j)$, according to proposition 3.2. Here the complex numbers $\alpha_j$, $\beta_j$ are obtained via the local exponents of the solutions of said equation around $z=0,\infty$. Also note, that $exp(2 \pi i \alpha_j)$, $j=1,...,n$ are the eigenvalues of $h_{\infty}$ and $exp(-2 \pi i \beta_j)$ are the eigenvalues of $h_0$ respectively.
I do understand, that if $d$ is of the form as in the remark, then the scalar shift is the effect on the monodromy group, by multiplying the solutions with $z^{-\delta}$. This can be calculated fairily easy, by using that as in (2.9) the solutions are of the form $z^{1-\beta_i} {}_n F_{n-1}(...)$. More precisely, by doing so we get local exponents $1-\beta_i-\delta$ in $z=0$ and $\alpha_i + \delta$ in $z=\infty$ respectively. Hence, we obtain a hypergeometric group $H(\tilde{a};\tilde{b})$ with parameters $$\tilde{a_i} = exp(2 \pi i(\alpha_i + \delta)) = exp(2 \pi i \delta)exp(2 \pi i \alpha_i) = d a_i,$$ $$\tilde{b_i} = exp(2 \pi i(\beta_i + \delta)) = exp(2 \pi i \delta)exp(2 \pi i \beta_i) = d b_i.$$
So as stated before, my remaining question now is:
Why are the associated reflection subgroups naturally isomorphic in this case?
*Edit: Bounty expires in one day!