= Solution
One example is \b[the identity function], $f(n)=n$. A second is the <permutation>
$$
\boxed{f(1)=2,\quad f(2)=3,\quad f(3)=1,\quad f(n)=n\ \text{otherwise}.}
$$
Both are <bijections> of $\mathbb N$, whichever convention includes or excludes zero. For the second, applying $f$ three times returns every input to itself. Thus $f^3=\operatorname{id}$, and composing with the <inverse function> gives $f^2=f^{-1}$. This <permutation cube identity criterion> also shows why three-cycles work whereas a nontrivial two-cycle would not.
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