Function with unique fixed point of all orders

Question

Let $f:[a,b]\to [a,b]$ be a continuous and monotone function. For each $n\ge 1$, there exists $x_n\in [a,b]$ such that $$f^n(x_n) = x_n,$$ where $f^n(x) = f(f(\cdots(f(x))\cdots))$ for $n$ times.

The function $f$ is nice if $x_n$ is unique for each $n$. For example, the function $f(x) = 1+x-\frac{x^2}{2}$ is nice on $[1,2]$, but $f(x) = 1+x-\frac{x^3}{3}$ is not nice in $\left[1,\frac{5}{3}\right]$.

My question is: how can we construct a nice function?

Thank you so much.

Answer 1

Let $f(x) = {1 \over 4}( 1+ 3x)$ on $[0,1]$.

Then $f^n(x) = x$ iff $x=1$.

It is straightforward to check that $f^n(x) = (1-({3 \over 4})^n) + ({3 \over 4})^nx$. Note that $(f^n)'(x) = ({3 \over 4})^n$, hence the fixed point is unique.