Are all vector-space valued functors on sets free? Let $\mathbf{Set}$ be the category of finite sets and functions between them, and let $\mathbf{Vect}$ be the category of finite-dimensional complex vector spaces and linear transformations between them. There is a free functor $G \colon \mathbf{Set} \to \mathbf{Vect}$, left adjoint to the forgetful functor, given by $G(X) = \mathbb{C}^X$ and $G(f)(\phi)(y) = \sum_{f(x)=y} \phi(x)$.

Is any functor $\mathbf{Set} \to \mathbf{Vect}$ of the form $H \circ G \circ F$ for $F \colon \mathbf{Set} \to \mathbf{Set}$ and $H \colon \mathbf{Vect} \to \mathbf{Vect}$?

A priori one might expect lots more functors, but I'm having a hard time coming up with any. On the other hand, functoriality seems to keep $G$ from "creating chaos" to "mess up freeness" (sorry that I can't explicate this feeling better). To keep it simple, let's keep things finite(-dimensional), and not consider anything about other base fields, or monoidal structure.
 A: There are some functors obtained by the "doubly contravariant trick" --- composites of the form $\mathbf{Set}\to\mathbf{Set}^{\mathrm{op}}\to\mathbf{Vect}$ or $\mathbf{Set}\to\mathbf{Vect}^{\mathrm{op}}\to\mathbf{Vect}$. I doubt all these can be represented as $H\circ G\circ F$...
A: The easiest example is the functor that sends the empty set to a 1-dimensional vector space, and all other finite sets to zero.  (Any H that sends some vector space to 0 must send 0 to 0 as well).
The second-easiest example is the one given by Jeremy Rickard.
If you're interested in functors from the category of finite sets to finite dimensional vector spaces, you might take a look at my paper http://arxiv.org/abs/1406.0786
The main result is a structure theorem about (finitely-generated) functors in this category.
A: This is probably an absurdly over-complicated answer, but ...
Let
$$J(X)=\left\{\sum_{x\in X}a_xx\in GX: \sum_{x\in X}a_x=0\right\}.$$
I claim that $J$ is not of the form $H\circ G\circ F$. 
Suppose it were.
Let $n=\{0,\dots,n-1\}$. 
The functors $F,G,H$ induce group homomorphisms
$$S(n)\to S(F(n))\to \operatorname{GL}(GF(n))\to\operatorname{GL}(HGF(n))$$
(where $S(n)$ is the symmetric group) and $J(n)$ is a representation (of dimension $n-1$) of all of these groups. It's irreducible as an $S(n)$-module, and so must be irreducible for all the other groups. (Actually, I'm really only going to need the case $n=3$.)
Since, for $0<n<m$, $n$ is a retract of $m$, and $J(n)\not\cong J(m)$, it follows that $F(n)\not\cong F(m)$ and $|F(1)|<|F(2)|<\dots$. So $\dim GF(n)\geq n-1 (=\dim HGF(n))$. But all homomorphisms $\operatorname{GL}(k,\mathbb{C})\to\operatorname{GL}(l,\mathbb{C})$ have abelian image for $k>l$ and so $\dim GF(n)=|F(n)|=n-1$ for $n>2$, or else $J(n)$ would be a direct sum of one-dimensional representations.
But $S(n-1)$ doesn't usually have any $(n-1)$-dimensional irreducible representations.
