A set $E$ is countable iff there is a bijection $f: E \to E$ that does not bijectively map nonempty proper subsets of $E$ to themselves

Question

Show that a set $E$ is countable if and only if there is a bijection $f: E \to E$, such that the only subsets $F$ of $E$ such that $f$ (when restricted to $F$) is a bijection from $F$ to $F$, are $F = \emptyset$ and $F = E$. (Countable includes finite here.)

Intermediate set theory, Drake and Singh

The "if" part is what I am wondering about.

Answer 1

Let $x\in E$ be any point, and consider the set $$F:=\{\dots,f^{-2}(x),f^{-1}(x),x,f(x),f^2(x),\dots\}$$