Is there a finite set, $M$, of functions $f\colon\mathbb{R}\to\mathbb{R}$ such that any polynomial $P\in\mathbb{Z}[x]$ can be expresed as a composition of functions from $M$?
For example I can do the following.
Let $\left \langle M \right \rangle$ be the set of all functions which can be expressed as a composition of functions from $M$. There are infinitely many primes $p_i$ such that: $p_ix = f_1\circ g_i\circ f_2$, where $f_1,f_2\in M$ and $g_i\in \left \langle M \right \rangle$. So $f_2$ is an injective function and $f_1$ is a surjective function...
Let $\alpha_0\colon\mathbb{R}\to (0,1)\subset\mathbb{R}$ be a bijection, then $\alpha_k=(x+1)^{\circ k}\circ\alpha_0\colon\mathbb{R}\to (k,k+1)\subset\mathbb{R}$ is also bijection. Let us somehow enumerate the set $\mathbb{Z}[x]=\{P_k |\ k\in\mathbb{N}_0\}.$
Define $f\colon\mathbb{R}\to\mathbb{R}$ as follows. $f\big|_{(k,k+1)}(x)=P_k\left(\alpha_k^{-1}(x)\right)$ and $f(x)=0$ for the remaining. Now it is quite easy to see that $$P_k=f\circ (x+1)^{\circ k}\circ\alpha_0.$$