我能想到最简解: f(f(1))=1. Assume f(1)=a. Considering f(f(f(1))) gives a=1.
Assume f(0)=b. Considering f(f(f(0))) leads to b^2-b+1=1. So b=0 or 1.
If b=0, then f(f(0))=0, contradiction. Hence f(0)=b=1.
我的Gemini 自己绕进去了,最后告诉我 “no continuous real-valued function” :)