Find all $a\in$ $\mathbb{R}$ such that there exists a bijection $f : [0,1] \rightarrow [0,1]$ satisfying $\forall x \in [0,1], f(f(x)+ax)=x$
This is most probably a contest math problem and therefore it's very hard (at least for me) to solve. Although, I have been thinking of ways to solve it but my incomplete knowledge in functional equations and the toughness of this problem are making by efforts worthless.
Any help is greatly appreciated.
The solution is only possible when $a=0$ (where $f(x)=x$).
Firstly, the equation can be written as: $$f^{-1}(x) - f(x) = ax$$
By contradiction, assuming $a>0$, then for all $x \in (0, 1]$ we have: $$f^{-1}(x) - f(x) = ax > 0$$ or $$ 0 ≤ f(x) < f^{-1}(x) ≤ 1$$
Since $f(x)$ is surjective, there exists an $x_0$ such that $f(x_0)=1$. If $x_0 \in (0, 1]$ it would violates the above inequalities, therefore $x_0=0$. But this implies $f^{-1}(1)=0$ which also violates the inequalities. We reached a contradiction.
The same argument applies to the case $a<0$, we conclude that $a=0$ is the only solution.