continuity and fixed point

Question

Show that a continuous mapping f from $[0,1]$ to $[0,1]$ which satisfies $f(f(x)) = x$ for all $x\in [0,1]$ and for which $f(x) \neq x$ for at least one $x\in [0,1]$ must have exactly one fixed point.

Answer 1

It is not true. The identity $f$ given by $f(x)=x$ for all $x\in[0,1]$ satisfies your assumption and has a continuum of fixed points.

Answer 2

Since $f(0)-0\geq0$ and $f(1)-1\leq0$, by the IVT, there must be at least one fixed point. Since $f$ is its own left-sided inverse, the function must be injective, hence it is either strictly increasing or strictly decreasing.

If it is decreasing, then everything is okay (indeed, the function $f(x)=-x+1$ has the properties, and even more generally, $f(x)=-x+(1-c)$). I claim that it cannot be increasing, therefore it must be decreasing. Since we know there is at least one $x$ where $f(x)\neq x$, we know either $f(x)>x$ or $f(x)<x$. Since strictly increasing functions preserve the inequality, we have $f\big(f(x)\big)>f(x)>x$ or vice versa, and in particular, the composition is not the identity. Thus, the function must be decreasing.

Now, since the function is strictly decreasing, it cannot have more than one fixed point, thus we have proven the conclusion.