$\mathbb{Q}$-automorphisms on $\mathbb{Q}(x_1, ..., x_n)$ map one-to-one into $S_n$

Question

Let $p(x) \in \mathbb{Q}[x]$ be a polynomial that doesn't have zeroes in $\mathbb{Q}$. Let $\mathcal{N} \in \mathbb{C}$ be the set of zeroes of $p(x)$. How to prove that the map

\begin{align} \mbox{Aut}_{\mathbb{Q}}(\mathbb{Q}(\mathcal{N})) &\to S_n \\ \phi \mapsto \phi \vert_{\mathcal{N}} \end{align}

is an injective group homomorphism?

The homomorphism part follows directly from the definition of the composition of functions, if I am not mistaken. But how to prove the injectivity part?