$$\newcommand\Ord{\text{Ord}}$$

Let $\mathbf{B}$ be the category of finite sets with bijections. We'll define two functors $\mathbf{B} \rightarrow \mathbf{Set}$. First, define $\Sym(\dash) : \mathbf{B} \rightarrow \mathbf{Set}$.

$$\Sym(X)<script type="math/tex">is the set of permutations of$$ X$. For a bijection$f : X \rightarrow Y$, define$\Sym(f) : \Sym(X) \rightarrow \Sym(Y)$by$\Sym(f)(\sigma) = f \circ \sigma \circ f^{-1}$$.

Now, define $\Ord(\dash) : \mathbf{B} \rightarrow \mathbf{Set}$.

$\Ord(X)$ is the set of total orders on $X$ -- for concreteness, let the elements of $\Ord(X)$ be relations in the sense of subsets of $X \times X$. Suppose $f : X \rightarrow Y$ is a bijection. Define $\Ord(f) : \Ord(X) \rightarrow \Ord(Y)$ by $\Ord(f)(R) = \setbuilder{(f(x), f(y))}{(x,y) \in R}$.

Theorem: There is no natural transformation $\Sym(\dash) \rightarrow \Ord(\dash)$.

Proof: Assume there were such a natural transformation $\alpha$. Let $A = \set{0, 1} \in \mathbf{B}$. And let $s$ be the swapping permutation $s(0) = 1$ and $s(1) = s(0)$. Then, we have a component $\alpha_A$ and by naturality $\alpha_A \circ \Sym(s) = \Ord(s) \circ \alpha_A : \Sym(A) \rightarrow \Ord(A)$. But, observe that:</script>

(\alpha_A \circ \Sym(s))(1_A) = \alpha_A(s \circ 1_A \circ s^{-1}) = \alpha_A(1_A)

$$which is some ordering, set $$\alpha_A(1_A) = R$$. Then,$$

(\Ord(s) \circ \alpha_A)(1_A) = \Ord(s)(R) = \setbuilder{(s(x), s(y))}{(x,y) \in R}

$which is clearly a different ordering than$R$$, a contradiction.

Thus, since $\Sym(X)$ and $\Ord(X)$ are the same size as sets, there is a bijection between them and we have $\Sym(X) \cong \Ord(X)$ in $\mathbf{Set}$ for each $X$, but not naturally. That is to say, $\Sym(\dash)$ and $\Ord(\dash)$ are not naturally isomorphic. This is a counterexample to the incorrect "converse" one might be willing to draw from exercise 1.3.26.