Bijection from $S_{n-1}$ to $\{\sigma \in S_{n} : \sigma(k) = j \}$ Let $n$ be a natural number.  Let $k$ be an element of $\{1, \ldots , n\}$.  For each j in $\{1, \ldots , n\}$, I want to find a bijection $f_j$ from $S_{n-1}$ to $\{\sigma \in S_n : \sigma(k) = j \}$, where $S_m$ denotes the symmetric group of order $m!$.  This bijection should (hopefully) have the property that for any $\sigma \in S_{n-1}$, $\text{sgn}(f_j(\sigma)) = (-1)^{j+k}\text{sgn}(\sigma)$.
I have attempted this several times and can't seem to find a bijection that works in every situation.  Also, I'm not sure how one would prove the property about the signs of the permutations.  Any help with one or both of these issues would be greatly appreciated.
 A: There is clearly a bijection
$$\tau:\{1,\cdots,n\}\backslash\{k\}\to \{1,\cdots,n-1\}$$
as they have the same number of elements. If $\sigma \in S_{n-1}$, we can create a permutation $\tilde{\sigma}$ from $S_n$:
$$\tilde{\sigma}(\ell)=\begin{cases}k & \ell = k \\ \tau^{-1}\sigma\tau(\ell) & \ell\ne k.\end{cases}$$
(We suppress the inclusion $\{1,\cdots,n\}\backslash\{k\}\hookrightarrow \{1,\cdots,n\}$ for brevity.)
The property you hope for doesn't hold: If $\sigma$ is the identity permutation, then $\rm sgn(\sigma)=sgn(\tilde{\sigma})=1$ and the choice of $k$ is irrelevant. Indeed, if $\rho$ is a transposition in $S_{n-1}$, then $\tilde{\rho}$ is likewise a transposition and so the parity is not altered at all by moving from $S_{n-1}$ into $S_n$. (Note that every possible embedding of the first into the second is obtained by some choice of $\tau$, so the failure of this property is not an artifact of an arbitrary choice.)
A: Regarding $S_{n-1}$ as the subgroup of $S_{n}$ which fixes $n$, let 
$f(\tau) = (kn)\tau (kn)(kj)$  if $k \not \in \{j,n\}$. If $k =j \neq n.$ just set $f(\tau) = (kn)\tau (kn).$  If $k = j = n$ there is nothing to do. If $k = n \neq j,$ 
set $f(\tau) = \tau (nj)$. Here, my permutations are acting from the right. It is clear that $\tau$ is a bijection in all cases. If $n >2,$ we can modify $f$ if we wish by multiplying the above choice on the right by a transposition which fixes $j.$ Hence when $n \geq 3,$ we can find both a sign preserving and a sign changing bijection. 
