Let $G =\langle A \rangle \leq S_n$ be a permutation group. The natural action of $G$ on $\Omega$ , where $|\Omega|=n$ is defined as follows:
$f:G \times \Omega \rightarrow \Omega$ such that $f(\sigma,\alpha)=\alpha^{\sigma}$ where $\alpha \in \Omega$ and $\sigma \in G$. The corresponding homomorphism is defined as follows:
$\phi : G \rightarrow Sym(n)$, $\sigma \rightarrow \phi ({\sigma})$, $\phi o {\sigma}(\alpha) =\alpha^{\sigma}$.
My question is the following:
Consider any normal subgroup $N$ of $G$. Then $N \leq S_n$ and $G/N $ is a group. What is the natural action of $G/N$ on $\Omega$ ? Can we say something even when $\Omega$ is some subgroup of $G$.
The action of a group $G$ on a set $\Omega$ factors through a quotient group $G/N$ if and only if $N$ is contained in the kernel of the morphism $\phi:G\to \mathrm{Sym}(\Omega)$, which is the intersection of all stabilizers of the elements of $\Omega$.
In your situation, the natural action of $G\leq S_n$ on $\Omega=\{1,\dots,n\}$ is faithful ($\ker\phi$ is trivial) so the action never factors unless $N$ is the trivial subgroup.