> > Or, to ask a simpler question to start with: is every finite category > concret(izabl)e? > Yes.
As a start, we know that every finite *group* is concretizable, that is: it is a subgroup of some permutation group.
Now we can make a similar argument for a finite category C; we will construct a faithful functor F:C-->FinSet. Given an object A, define F(A) to be the set of all morphisms in C with codomain A. Given a morphism f:A-->B, define F(f) to be the function F(A)-->F(B) that is just composition with f.
Now if f,g:A-->B are two different morphisms, the funtions F(f) and F(g) are also different: evaluate the functions at the identity of A:
F(f)(id_A) = f*id_A = f F(g)(id_A) = g*id_A = g
So F is indeed faithful.
Clearly this argument is just an elaboration of the finite group statement.