Let $\Phi$ be a root system in $E=\mathbb{R}^l$, $\Delta=\{\alpha_1,\cdots,\alpha_l\}$ be a base of $\Phi$.
Let $\langle\alpha,\beta\rangle:=\frac{2(\alpha,\beta)}{(\beta,\beta)}$ for $\alpha,\beta\in\Phi$.
Let $\Lambda:=\{\lambda\in E \, | \, \langle \lambda,\alpha_i\rangle\in\mathbb{Z} \,\,\forall i=1,2,\cdots,n\}.$
Then $\Lambda$ is a lattice in $E$ of rank $l$, and contains the sublattice $\Lambda_r:=\mathbb{Z}\alpha_1 + \cdots + \mathbb{Z}\alpha_l$.
Let $\sigma$ be an automorphism of root system $\Phi$, i.e. $\sigma:E\rightarrow E$ is a linear isomorphism such that $$\langle \sigma(\alpha),\sigma(\beta)\rangle=\langle\alpha,\beta\rangle \,\,\forall\alpha,\beta\in\Phi.$$ It can be shown that the group ${\rm Aut}\Phi$ is semi-direct product of $W$ and $\Gamma$, where $W$ is Weyl group and $\Gamma=\{\sigma\in{\rm Aut}\Phi\,|\,\sigma(\Delta)=\Delta.\}$
Further it can be shown easily that for $\sigma\in{\rm Aut}\Phi$, $\sigma(\Lambda)\subset\Lambda$ and $\sigma(\Lambda_r)\subset\Lambda_r$. Hence $\sigma$ induces action on $\Lambda/\Lambda_r$.
Question: If $\sigma\in\Gamma$ is non-identity, then its action on $\Lambda/\Lambda_r$ is non-trivial; why?
This is an exercise in Humphreys' Lie algebra, Section 13.
My way: I did computational proof, namely I considered those irreducible systems $\Phi$ for which $\Gamma$ is non-trivial (for example $A_l$ with $l\ge 2$, $D_l$ etc.); in this case, $\Gamma$ is well-known (see a table in Section 13 of Humphreys' Lie algebra). There is a set of coset representatives of $\Lambda_r$ in $\Lambda$ given in a table in the book of Humphreys. Then the non-trivial element of $\Gamma$ can be seen to act non-trivially on some coset $\lambda_i+\Lambda_r$. So this is explicit case-by-case proof of statement.
But, I couldn't proceed to the theoretical proof for statement in question.
Any hint?
It seems the statement in the exercise is not true as stated. For example consider $\Phi = G_2\times G_2$, then $\text{Aut }\Phi/W\cong S_2$, while $\Lambda/\Lambda_r$ is trivial. Probably the statement was meant for $\Phi$ to be irreducible, in which case I guess the most straightforward way would be a case-by-case computation as you did.