This was one of the challenging permutation problems in a discrete maths textbook that I couldn't solve.
Question: Let $C_n = \{\pm1,\pm2,\pm3,...,\pm n \}.$ How many permutations $f : C_n\rightarrow C_n$ are there such that $f(-x) = -f(x)$ for all $x \in C_n$?
Answer: $2^n \times n!$ or $(2^n)(n!)$
I've tried to draw arrow diagrams to visualize small examples of the scenario described in the problem but I still can't see why there are $(2^n)(n!)$ possibilities.
Once you define what $f(x)$ for $1\le x\le n$ is, the rest of $f$ is completely determined by $f(-x)=-f(x)$. Since there are only two elements with a given magnitude in $C_n$, the magnitudes of $f(1),\dots,f(n)$ must all be distinct, otherwise there would be a pair of numbers in $C_n$ that $f$ maps to the same value and $f$ would not be a permutation. Thus the magnitudes of $f(1),\dots,f(n)$ are a permutation of $1,\dots,n$, and we can choose their signs independently, giving the number of functions $f$ as $2^nn!$.