[Tom Leinster, Exercise 1.3.31] A permutation of a set $X$ is a bijection $X \rightarrow X$. Write Sym$(X)$ for the set of permutations of $X$. A total order on a set $X$ is an order $\leq$ such that for all $x, y \in X$, either $x \leq y$ or $y \leq x$; so a total order on a finite set amounts to a way of placing its elements in sequence. Write Ord$(X)$ for the set of total orders on $X$. Let $\mathscr{B}$ denote the category of finite sets and bijections. (a) Give a definition of Sym on maps in $\mathscr{B}$ in such a way that Sym becomes a functor $\mathscr{B} \rightarrow$ Set. Do the same for Ord. Both your definitions should be canonical (no arbitrary choices). (b) Show that there is no natural transformation Sym $\rightarrow$ Ord. (Hint: consider identity permutations.) (c) For an $n$-element set $X$, how many elements do the sets Sym$(X)$ and Ord$(X)$ have? Conclude that Sym$(X) \cong Ord(X)$ for all $X \in \mathscr{B}$, but not naturally in $X \in \mathscr{B}$. (The moral is that for each finite set $X$, there are exactly as many permutations of $X$ as there are total orders on $X$, but there is no natural way of matching them up.)
I am aware of the similar question, but I solved it in this way...
Since $\mathscr{B}$ is category of finite sets and bijections, so if $X, Y \in \mathscr{B}$ such that there is a map $ f: X \rightarrow Y$, since $f$ is a bijection and $X$ and $Y$ are finite sets, hence $|X| = |Y|$. So Sym$(X)$ $=$ Sym$(Y)$ $\in$ Set, where Set is the category of sets. Hence, I think $(?)$ that Sym$(f)$ is nothing but identity morphism on Sym$(X)$. But I can't understand what Ord$(f)$ should be. I know that $|$ Ord $(X)$ $|$ $=$ $n!$, if $|X| = n$, but I don't know how to proceed. Also, I don't understand how to solve the part (b), even using the hint.
It will be beneficial if someone helps me to solve the problem and clarify my doubts.
Thanks in advance!