Reference: Page 140 of the book in the title.
I know this question may be difficult to answer on here, but regarding $\mathbb{R}^{n}$ as a subset of the nth Clifford algebra $Cl_{n}$ for each $u \in \Gamma_{n}$, the function $\rho_{u}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ defined by $\rho_u(v)=\alpha(u)vu^{-1}$ is an $\mathbb{R}$-linear isometry, where $\alpha: Cl_{n} \rightarrow Cl_{n}$ is the canonical automorphism and $\Gamma_{n}$ is the nth Clifford group.
I don't comprehend the second equality in his proof given. He says
For any $x \in \mathbb{R}^{n}$, $|\rho_{u}(x)|^{2}=-\rho_{u}(x)^{2}=-((-u)xu^{-1})^{2}$.
How do we get the second equality? If we write out $-\rho_{u}(x)^2=-( \alpha(u)xu^{-1})^{2}$ it appears he's claiming $\alpha(u)=-u$, but I don't think this is true for every $u \in \Gamma_{n}$.