Let $R$ be a root system and $W$ be the Weyl group of $R$. Then we have that the group $W$ is a normal subgroup of the group $Aut(R)$ of automorphisms of $V$ ($V$ a finite dimensional vector space) leaving $R$ invariant. This should follow from the following lemma:
Let $\phi \in Aut(R)$ and $\alpha \in R$. Then applies $$s_{\phi(\beta)}=\phi \circ s_{\beta} \circ \phi^{-1}$$
The proof should follow directly from this lemma: Let $\alpha$ be a nonzero element of $V$, and let $R$ be a finite subset of $V$ which spans $V$. There is at most one symmetry with vector $\alpha$ which leaves $R$ invariant.
I don't see the connection between these two lemma. How can I conclude that the above statement is true? Can someone please help me?
Any element of $\mathcal{W}$ can be written $\sigma_{\alpha_1}\cdots\sigma_{\alpha_r}$ for $\alpha_i\in R$. Then for $\tau\in{\rm Aut}(R)$, we have \begin{equation*} \tau(\sigma_{\alpha_1}\cdots\sigma_{\alpha_r})\tau^{-1} = (\tau\sigma_{\alpha_1}\tau^{-1})\cdots(\tau\sigma_{\alpha_r}\tau^{-1}) =\sigma_{\tau(\alpha_1)}\cdots\sigma_{\tau(\alpha_r)} \in\mathcal{W} ; \end{equation*} where the last equality is Lemma 9.2. in Humphreys.