This is probably an absurdly over-complicated answer, but ...

Let $$A(X)=\left\{\sum_{x\in X}a_xx\in GX: \sum_{x\in X}a_x=0\right\}.$$

I claim that $A$ 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 $A(n)$ 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 $A(n)\not\cong A(m)$, 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 $A(n)$ would be a direct sum of one-dimensional representations.

But $S(n-1)$ doesn't usually have any $(n-1)$-dimensional irreducible representations.

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.