tchebytchev wrote: Find all finite and non-empty sets $A$ of functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that for all $f_1, f_2 \in A$, there exists $g \in A$ such that for all $x, y \in \mathbb{R}$ $$f_1 \left(f_2 (y)-x\right)+2x=g(x+y)$$ Setting $x=0$, we get $g(x)=f_1(f_2(x))$ and so composition of functions is internal to $A$ and assertion is : Assertion $P(f_1,f_2,x,y)$ : $f_1(f_2(y)-x)+2x=f_1(f_2(x+y))$ Let $f\in A$. $P(f,f,f(x)-x,x)$ $\implies$ $f(f(f(x)))=3f(x)-2x$ and so $3f-2Id\in A$ $P(f_1,f_2,x,0)$ $\implies$ (1) : $f_1(f_2(0)-x)+2x=f_1(f_2(x))$ $P(3f_1-2Id,f_2,x,0)$ $\implies$ (2) : $3f_1(f_2(0)-x)-2(f_2(0)-x)+2x=3f_1(f_2(x))-2f_2(x)$ 3(1)-(2) $\implies$ $f_2(x)=x+f_2(0)$ And so (stable by composition) $x+kf_2(0)\in A$ and so $f_2(0)=0$ since $A$ is finite And so $\boxed{A=\{\text{Identity}\}}$ which indeed is a solution If we remove "finite" constraint