Problem 1178

Solution: To begin, we show that $A$ is a set of surjective functions. Indeed, for $f \in A$ and $f_{1}=f_{2}=f$, with $x=-y$,

$$f(f(y)+y)=g(0)+2 y$$

and thus $f$ is surjective.

Furthermore, with $x=0$, we obtain that if $f_{1}, f_{2} \in A$ then $f_{1} \circ f_{2} \in A$.

Since $A$ is non-empty, we take $f \in A$. By the previous remark, we know that $f^{(k)} \in A$ for all $k \in \mathbb{N}$. As $A$ is finite, there exist $k_{1}>k_{2}$ such that $f^{\left(k_{1}\right)}=f^{\left(k_{2}\right)}$. Since $f^{\left(k_{2}\right)} \in A$, $f^{\left(k_{2}\right)}$ is surjective and thus with $y=f^{\left(k_{2}\right)}$, we obtain

$$f^{\left(k_{1}-k_{2}\right)}(y)=y, \forall y \in \mathbb{R}$$

Since $f^{\left(k_{1}-k_{2}\right)} \in A$, $A$ contains the identity. It is easily verified that $A=\{i d\}$ is a possible set. We now show that $A$ cannot contain anything other than the identity.

By setting $x=0$ in our condition, we obtain that $g=f_{1} \circ f_{2}$ and therefore for all $f_{1}, f_{2} \in A$

$$f_{1}\left(f_{2}(y)-x\right)+2 x=f_{1}\left(f_{2}(x+y)\right)$$

If $A$ contained a function $h$ other than the identity, with $f_{1}=i d, f_{2}=h$ and $y=0$, we would have

$$h(x)=x+h(0) \text { and thus } h^{(n)}(x)=x+n h(0)$$

Since $A$ is finite and $h^{(n)} \in A$ for all $n \in \mathbb{N}$, we have $h(0)=0$ and thus $h \equiv i d$.