I have to show that ${\rm Aut}(C)$ is a subgroup of $S_{n}$.
I know that if $C$ is a code of length $n$, then the automorphism group (${\rm Aut}(C)$) of $C$ are the permutations of $n$, with the property that if we apply the corresponding permutation of coordinates in a codeword of $C$, we take a (maybe different) codeword of $C$.
If I remember correctly, the automorphism group of a code is defined as the set$$ \operatorname{Aut}(\mathcal{C})=\{\sigma\in S_n:\sigma(\mathcal{C})=\mathcal{C}\}\subseteq S_n. $$
We prove the group axioms: Let $\sigma_1,\sigma_2\in S_n$. Then we have$$ \sigma_1(\mathcal{C})=\mathcal{C}\text{ and }\sigma_2(\mathcal{C})=\mathcal{C}.$$We obtain$$ (\sigma_1\circ\sigma_2)(\mathcal{C})=\sigma_1(\sigma_2(\mathcal{C}))=\sigma_1(\mathcal{C})=\mathcal{C}. $$Therefore the set is closed under composition. The identity permutation is clearly an element of $\operatorname{Aut}(\mathcal{C})$. We only have to show that the inverse $\sigma_1^{-1}$ is an element. But we have$$ \sigma_1^{-1}(\mathcal{C})=\sigma_1^{-1}(\sigma_1(\mathcal{C}))=(\sigma_1^{-1}\circ\sigma_1)(\mathcal{C})=(\operatorname{id})(\mathcal{C})=\mathcal{C}. $$ We get $\sigma_1^{-1}\in\operatorname{Aut}(\mathcal{C})$. So $\operatorname{Aut}(\mathcal{C})$ satisfy the group axioms and therefore is a subgroup of $S_n$.