Find $f,g:[1,2]\to[1,2]$ such that $f\circ g=g\circ f=$constant but for no $x\in[1,2]$ we have $f(x)=g(x)=x$

Question

Does there exist mappings $f,g:[1,2]\to[1,2]$ such that $f\circ g=g\circ f=$constant but for no $x\in[1,2]$ we have $f(x)=g(x)=x?$

Linked: $f,g:[1,2]\to[1,2]$ such that $fg=gf$ but for no $x\in[1,2]$ we have $f(x)=g(x)=x$

Answer 1

Let $f(g(x))=g(f(x))=c$ for some $c\in [1,2]$ and all $x\in[0,1]$. Therefore $$f(c)=f(g(f(1))=c$$ and $$g(c)=g(f(g(1)))=c$$ So the pair of functions that you seek do not exist.