Bijective Proof: $\sum_{k=0}^n \binom{n}{k} (2k-1)!! (2n-2k-1)!!=2^n \cdot n!$ This is problem 54 in Stanley's Bijective Proof Problems: 

If $X \subseteq \mathbb{P}$, then write $-X=\{-n: n ∈ X\}$. Let $g(n)$ be the number of ways to choose a subset $X$ of $[n]$, and then choose fixed point free involutions $\pi$ on $X \cup (-X)$ and $\overline{\pi}$ on $\overline{X} \cup (-\overline{X})$, where $\overline{X}=\{i ∈ [n]: i \not ∈ X\}$. Then $g(n) = 2^n n!$.

I was able to prove this by using exponential generating functions, and found another proof by using some algebraic manipulations to reduce the problem to showing that $$\sum_{n=0}^k \binom{2n-2k}{n-k}\binom{2k}{k}=4^n,$$ which is a question that has been asked before on this website (for example, here). 
However, I'm unable to produce a bijective proof of this. I'm not seeking a full solution, as this is a homework problem, but I'm feeling very stumped about finding a combinatorial object to relate to the fixed-point-free involution interpretation of $g(n)$, and would appreciate a pointer in the right direction.
I tried to view $2^n \cdot n!$ as the number of length $n$ walks to to a corner of the unit cube in $\mathbb{R}^n$ (taking one step parallel to each coordinate axis $\vec{e_j}$ and choosing whether to step $+\vec{e}_j$ or $- \vec{e}_j$), but couldn't relate this to the construction "choose a subset $X \subseteq [n]$, and then fixed-point-free involutions $\pi$ and $\overline{\pi}$."
I also noticed that $$g(n) \cdot \underbrace{(2n-1)!!}_{=\#\{\text{fixed-point free involutions of $[2n]$} \}}=(2n)!,$$ and wondered if the orbit-stabilizer theorem might somehow be applied, but again could not find a good construction. 
 A: HINT as requested
A fixed-point-free involution is of course just a set of unordered-pairs $\{\dots, \{a, \pi(a)\}, \dots\}$.  The interesting thing about an involution on $X \cup (-X)$ is that it has a natural many-to-one mapping to a permutation on $X$.  E.g. Suppose $X = \{1,2,5,6\}, \pi = \{\{1, -1\}, \{2, 5\}, \{-5, -6\}, \{6, -2\}\}$.  If you ignore the $\pm$ signs, this naturally maps to the permutation $\sigma = {{1\ 2\ 5\ 6} \choose {1\ 5\ 6\ 2}}$ (two line notation) or equivalently $(1)(256)$ (cycle notation).  I.e. start with a number $x$, find its partner $\pi(x)$, then find $-\pi(x)$, find its partner $\pi(-\pi(x))$, etc.  Eventually you must close a cycle and end at $-x$.
Now, there are two problems:
First, note that $\{\{1, -1\}, \{2, \pm5\}, \{\mp5, \pm6\}, \{\mp6, -2\}\}$ maps to the same $\sigma$.  However you can encode the $\pm$ signs if only you could "decorate" the cycle with several bits.
Second, remember each pair is unordered.  So we need a method to make sure we don't read the same $\pi$ as $\{\{1, -1\}, \{-2, 6\}, \{-6, -5\}, \{5, 2\}\}$ and then map it to $(1)(265) \neq \sigma$.


*

*For a cycle of length $k$, how many bits do you need to encode all the $\pm$ signs, and to make sure your encoding is good enough to not confuse the cycle $(256)$ vs the cycle $(652)$?

*The given answer of $2^n n!$ gives you $n$ bits to decorate a permutation on $[n]$.  Can you find a way to use them all to encode $(\pi, \bar{\pi})$?
Lemme know if you need more hints.
