Let $f \colon [ 0, 1] \to [0, 1]$ be a continuous map such that $$ f\big( f(x) \big) = x \ \mbox{ for each } x \in [0, 1], $$ and $$ f(x) \neq x \ \mbox{ for at least one } x \in [0, 1], $$ then how to show that $f$ has exactly one fixed point in $[0, 1]$?
I know (how to show) that $f$ does have at least one fixed point in $[0, 1]$.
Suppose that $u \in [0, 1]$ is such that $f(u) \neq u$. Suppose further that $p, q \in [0, 1]$ satisfy $f(p) = p$ and $f(q) = q$.
What next? Can we arrive at a contradiction?
Of course, $$ f\big( f(u) \big) = u, $$ so that $$ f \big( f(u) \big) \neq f(u). $$
Context:
This is Prob. 1.2 in the book An Introduction To Metric Spaces And Fixed Point Theory by Mohamed A. Khamsi and William A. Kirk.