Continuous functions that are periodic under composition

Detailed Solution

Since $f^n(x)=x$ for all $x$, $f$ is injective (if $f(x)=f(y)$ then $f^n(x)=f^n(y)$, i.e. $x=y$) and surjective (every $y$ has preimage $f^{n-1}(y)$, since $f(f^{n-1}(y))=f^n(y)=y$). A continuous injective function on an interval must be strictly monotone, so $f$ is strictly monotone on $[0,1]$ — this is (D). (As a consequence, $f$ also has at least one fixed point: if increasing, the bijection forces $f(0)=0$; if decreasing, it forces $f(0)=1,f(1)=0$, and the Intermediate Value Theorem applied to $f(x)-x$ gives a root — so (B) is also true, but (D) is the more fundamental fact.) Option (A) is false: $f(x)=1-x$ satisfies $f^2(x)=x$ (so $n=2$ works) yet is not the identity. Option (C) is false: for that same $f(x)=1-x$, the point $x=1/2$ satisfies $f(1/2)=1/2$, so it has period $1$, not period exactly $2$. Thus the correct option is $\boxed{\mathrm{D}}$.