I am checking some basic properties of Clifford multiplication while reading the book "Twistors and Killing spinors on Riemannian manifolds" by Baum et al. It is said that Clifford multiplication is invariant under the action of $\mathrm{Spin}(n)$, and this is where I am stuck.
Clifford multiplication is a particular product between a vector $x \in \mathbb R^n$ and a spinor $\varphi \in \Delta_n$, where $\Delta_n$ is the $\mathrm{Spin}(n)$-representation obtained by restriction of the standard representation $\rho$ of the Clifford algebra $Cl(\mathbb R^n) \otimes \mathbb C$. I should discuss a bit more carefully this point, but I do not think it will be relevant for my question (maybe I am wrong). By definition, viewing $x \in \mathbb R^n$ in $Cl(\mathbb R^n)$, we have $$x \cdot \varphi := \rho(x)\varphi.$$ Now $\mathrm{Spin}(n)$ also acts on $\mathbb R^n$. In fact the map $\lambda\colon \mathrm{Spin}(n) \to \mathrm{SO}(n)$ such that $\lambda(u)x = uxu^{-1}$ is the double covering of $\mathrm{SO}(n)$, and here $x$ is again a vector in $\mathbb R^n$.
The problem now is to prove that Clifford multiplication is $\mathrm{Spin}(n)$-invariant, where $\mathrm{Spin}(n)$ acts on $\mathbb{R}^n$ via $\lambda$ and on $\Delta_n$ via $\rho$. Given $u \in \mathrm{Spin}(n)$ this should amount to saying that $\lambda(u)x \cdot \rho(u)\varphi = x\cdot \varphi$. However, what I get is \begin{align*} \lambda(u)x \cdot \rho(u)\varphi & = uxu^{-1}\cdot \rho(u)\varphi \\ & = \rho(uxu^{-1})\rho(u)\varphi \\ & = \rho(uxu^{-1}u)\varphi \\ & = \rho(ux)\varphi \\ & = \rho(u)\rho(x)\varphi = \rho(u)x \cdot \varphi. \end{align*} This seems to prove that Clifford multiplication is $\mathrm{Spin}(n)$-equivariant, not invariant. Does anybody see where I am wrong?