$\mathbf C\mathbf a\mathbf t$ as a concrete category Consider the category Cat as a concrete category over Set $\times$ Set via the functor 
U : Cat $\rightarrow$ Set $\times$ Set, defined by
U$(\mathbf A \xrightarrow{F} \mathbf B) = ($Ob$(\mathbf A)\xrightarrow{F_O}$ Ob$(\mathbf B)$ , Mor$(\mathbf A)\xrightarrow{F_M}$ Mor$(\mathbf B))$, 
where $F_O$ is the restriction of $F$ to objects and $F_M$ is its restrictions to morphisms.
Cat is the category of all small categories in the sense that their objects and morphisms form sets (not classes).
It is clear that this concrete category is transportable. But is this category uniquely transportable?

A concrete category $(\mathbf A, U)$ over $\mathbf X$ is said to be (uniquely) transportable provided that for every $\mathbf A-$object A and every $\mathbf X-$isomorphism $UA\xrightarrow{k}X$ there exists a (unique) $\mathbf A-$object B with $UB=X$ such that $A\xrightarrow{k}B$ is an $\mathbf A-$isomorphism.
 A: Let $X = (O, M)$ be a pair of sets, and let $A$ be a category with set of objects $O_A$ and set of morphisms $M_A$. Suppose given a pair $k = (k_0, k_1)$ of isomorphisms $k_0: O_A \to O$, $k_1: M_A \to M$. Let $s_A, t_A: M_A \to O_A$ be the source and target functions for $A$; then the only way to define the source and target functions $s, t: M \to O$ on the lifted structure over $X$ in such a way that $k$ preserves source and target data is by setting 
$$s = k_0 \circ s_A \circ k_1^{-1}, \qquad t = k_0 \circ t_A \circ k_1^{-1}$$ 
(that's what I meant by "conjugating" in my comment). 
It's the same idea to define the lifted structure $i, c$ for identity and composition. If $i_A, c_A$ are the identity and composition on $A$ and $k$ is to preserve identity and composition data, then there's only one way to define them: put 
$$i = k_1 \circ i_A \circ k_0^{-1}, \qquad c = [(f, g) \mapsto (k_1 \circ c_A)(k_1^{-1} f, k_1^{-1} g)]$$ 
where $f, g$ is any composable pair in the lifted source-target structure over $X$. 
Of course unique transport of structure via conjugation is a simple idea which applies much more generally, as hinted in Peter's second comment. 
