关于Tom Leinster习题1.3.31中有限集双射范畴函子及自然变换问题的求助

[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$(X) \cong \text{Ord}(X)$ 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 a 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!

备注：内容来源于stack exchange，提问作者delusional.existence