We know the symmetric group is defined over any set is the group whose elements are all the bijections from the set to itself.
S$_n$ is defined over a finite set of $n$ symbols consists of the permutation operations that can be performed on the $n$ symbols.
Suppose these $n$ symbols are:
$$(1,2,3,\dots,n)$$
other than any of the permutation of order $n!$.
Let us also allow an additional $\pm$ sign as a $\mathbb{Z}/2\mathbb{Z}$ cyclic group on each symbols
$$\pm 1,\pm 2, \pm 3,\dots,\pm n$$
We allow all operations including the permutation and the $\mathbb{Z}/2\mathbb{Z}$ cyclic group on each symbol.
Let us call this the generalized symmetric group ${\bf S'}_n$, which has an order
$$(2^n) n!$$
So a generic group element can be represented by
$$ (1,2,3,\dots,n) \to (2,-3,\dots,-(n-k),k) $$ or anything now you can imagine.
Question 1: How is this group related to another more familiar group, say a group extension from $S_n$?
A partial answer* For $n=2$, we have the order-2 group $$S_2=\mathbb{Z}/2\mathbb{Z}$$ and the order-8 group $${\bf S'}_2=D_8$$ So the answer I look for $n=2$ is that, the $$ 0 \to \mathbb{Z}/2\mathbb{Z} \to {\bf S'}_2=D_8 \to (\mathbb{Z}/2\mathbb{Z})^2 \to 0 $$ $$ 0 \to \mathbb{Z}/4\mathbb{Z} \to {\bf S'}_2=D_8 \to \mathbb{Z}/2\mathbb{Z} \to 0 $$ $$ 0 \to (\mathbb{Z}/2\mathbb{Z})^2 \to {\bf S'}_2=D_8 \to \mathbb{Z}/2\mathbb{Z} \to 0 $$ There are several $ \mathbb{Z}/2\mathbb{Z}$ above as quotient group or normal subgroup of ${\bf S'}_2=D_8$,
but none of normal subgroup is directly the $S_2=\mathbb{Z}/2\mathbb{Z}$. Yes or no? But we have $$ 0 \to (\mathbb{Z}/2\mathbb{Z})^2 \to {\bf S'}_2=D_8 \to {S}_2=\mathbb{Z}/2\mathbb{Z} \to 0 $$ for a good choice of a normal subgroup $(\mathbb{Z}/2\mathbb{Z})^2$.
Question 2: Do we have a similar structure for $$ 0 \to ? \to {\bf S'}_n \to ? \to 0 $$ Can we find $S_n$ sitting somewhere in ${\bf S'}_n$ related to quotient group or normal subgroup of ${\bf S'}_n$?
Maybe also be related: https://groupprops.subwiki.org/wiki/Generalized_symmetric_group