Question on a permutation of the roots of a Galois number field, dihedral over the rationals The polynomial $f(x) = x^6 - 7x^5 + 21x^4 - 41x^3 + 63x^2 - 63x + 27$ defines a Galois extension $H$ of $\mathbb{Q}$. The Galois group of the extension $H/\mathbb{Q}$ is dihedral, and depending on your notation is $D_6$ or $D_3$.
(Aside: in fact I generated the polynomial by defining $g_1(\tau) = \eta(\tau/3)/\eta(\tau)$, where $\eta(\tau)$ is the Dedekind eta function, and letting $w = g_1(\tau)^2$ where $\tau = (-1 + \sqrt{-23})/2$ and taking the minimum polynomial of $w$ over $\mathbb{Q}$. In fact $w$ generates the Hilbert class field $H$ of $\mathbb{Q}(\sqrt{-23})$ over $\mathbb{Q}$.)
Suppose that the roots of $f(x)$ are written $r_1, r_2, \ldots, r_6$. Also suppose that they are ordered such that $r_1, r_2$ and $r_3, r_4$ and $r_5, r_6$ are complex conjugate pairs.
An automorphism in $G =$ Gal$(H/\mathbb{Q})$ defines a permutation on the roots.
My question is this: is there a $\sigma \in G$ which as a permutation on the roots is given by $(r_1 r_2)(r_3 r_4)(r_5 r_6)$.

I present here two different answers. The first is how I see it. The second is how another mathematicians sees it, even after I explained my answer and followed it with a four hour discussion. I wonder if anyone else is convinced by the second answer?
Answer 1: Let $\tau$ be complex conjugation. It is well known and easy to see that $\tau(H) = H$. Clearly $\tau$ fixes $\mathbb{Q}$ and thus it is a $\mathbb{Q}$-automorphism of $H$. By definition it is therefore an element of Gal$(H/\mathbb{Q})$ and by definition it has the required action on the roots of $f$.
Answer 2: If complex conjugation were a Galois automorphism of $H$ then $H$ would have a totally real subfield. As it does not (admitted), it is not a Galois automorphism.
$H$ is the field $\mathbb{Q}[x]/(f)$. Given an ordering of the roots $r_1, \ldots, r_6$ there is a map $\phi : H \to \mathbb{C}^6$ which is defined by $x \pmod{f} \mapsto (r_1, r_2, \ldots, r_6)$.
An automorphism in $G =$ Gal$(H/\mathbb{Q})$ induces a permutation on the roots. However, there is no automorphism in $G$ which maps $(r_1, r_2, r_3, r_4, r_5, r_6)$ to $(r_2, r_1, r_4, r_3, r_6, r_5)$. Instead an automorphism that permutes the roots so that $r_1$ is sent to $r_2$ and vice versa will send $(r_1, r_2, r_3, r_4, r_5, r_6)$ to something like $(r_2, r_1, r_6, r_5, r_4, r_3)$.
A demonstration follows using the favourite computer algebra system of this mathematician, which confirms his claim.
Are there two fundamentally different definitions of the terms used in my question in the literature? Specifically, what is meant by a permutation of the roots given by a Galois automorphism?
