Is there a slick argument to prove that, for $n>6$, $2(n-2)!=2^kk!(n-2k)! \implies k=1$? In a book, this implication is used to prove that every automorphism of $S_n$ is interior if $n \neq 6$ by comparing the cardinality of the centralizer of a transposition and the centralizer of the image of said transposition by an automorphism (which has to be a product of $k$ transpositions due to it being of order $2$). The equality $k=1$ shows an automorphism of $S_n$ for $n\neq 6$ sends transpositions on transpositions and permits to conclude that it is interior. However, it is not obvious to me that $k=1$...
My first idea is obviously to try to compare the exponent of $2$ in the prime decomposition, but this doesn't help too much
$$1+\sum_{\ell=1}^\infty\left\lfloor \frac{(n-2)}{2^\ell}\right\rfloor = k+\sum_{\ell=1}^\infty\left\lfloor \frac{k}{2^\ell}\right\rfloor +\sum_{\ell=1}^\infty\left\lfloor\frac{n-2k}{2^\ell}\right\rfloor$$
Is there a slick way to go about it without using a computer? The only thing that's obvious to me is that if $k>1$, then one side grows much larger than the other side for $n$ large enough, so if $n$ is large enough then $k=1$, but I don't see how "large enough" means $n>6$. It is obvious that $k$ is odd so if $k>1$ then $k \geq 3$ so it is intuitive that the "large enough" is not so large, but I'd like an easy argument to prove that $n>6$ shall suffice.
 A: Let $N=n-2$ and $K=k-1$ so your equality becomes $N!=2^K(K+1)!(N-2K)!$ or equivalently $$\binom N{2K}=\frac{2^K(K+1)!}{(2K)!}.$$ Over the positive integers, the RHS is strictly decreasing and is less than $1$ for all $K>2$. When $K=1$, the equality becomes $\binom N2=2$ and when $K=2$, the equality becomes $\binom N4=1$. Both equations contradict $N>4$, forcing $K=0$ (where both sides equal $1$).
A: This is perhpas a more intuitive approach. If we count the number of factors on both sides that are not $1$, the left side has $1 + (n-3) = n-2$ while the right also has $k + (k-1) + (n-2k-1) = n-2$.
Now we can match every factor on the left to a smaller or equal one on the right:
The left factors are 2, 2, 3, 4, ..., n-2, where there are exactly two 2s. Unless $k=1$, the right will have too many $2$'s that make it smaller than the left product.
A: We have that for $k=1$
$$2^kk!(n-2k)!=2(n-2)!$$
For $k=2$ by inspection we have
$$ 2^kk!(n-2k)!=4\cdot 2!(n-4)!<2(n-2)! \iff
   (n-2)(n-3)>2\cdot 2!$$
which is true since $(n-2)(n-3)>12$.
For $k=3$
$$2^kk!(n-2k)!=8\cdot 3!(n-6)!<2(n-2)! \iff
   (n-2)(n-3)(n-4)(n-5)>4\cdot 3!$$
which is true since $(n-2)(n-3)(n-4)(n-5)>4!$.
More in general for $k>3$
$$ 2^kk!(n-2k)!<2(n-2)! \iff \overbrace{(n-2)\cdots
   (n-2k+1)}^{2k-2 \, \text{terms}}>2^{k-1}\cdot k!$$
which is true since $\overbrace{(n-2)\cdots
   (n-2k+1)}^{2k-2 \, \text{terms}}>(2k-2)!>2^{k-1}\cdot k!$, indeed
$$\frac{(2k-2)!}{k!}=\overbrace{(2k-2) \cdots (k+1)}^{k-2 \, \text{terms}}>(k+1)^{k-2}>2^{k-1}$$
since
$$(k+1)^{k-2}>2^{k-1}\iff \frac12 \left(\frac{k+1}2\right)^{k-2}>1$$
which is true for $k>3$.
