Let $n_1,n_2,\ldots$ be the increasing sequence of odd numbers $\ge 1$, put, for all $k\ge 0$, $f(n_{1989k+i})=n_{1989k+i+1},\ \forall i\in\overline{1,1988}$, $f(n_{1989k+1989})=2n_{1989k+1}$, and then extend the function to even numbers by using $f(2n)=2f(n),\ \forall n\ge 1$.

orl wrote: Let $ \mathbb{N} = \{1,2, \ldots\}.$ Does there exists a function $ f: \mathbb{N} \mapsto \mathbb{N}$ such that $ \forall n \in \mathbb{N},$ $ f^{1989}(n) = 2 \cdot n$ ? Define for all $ n \in \mathbb{N}$: $ m_n = \frac {\frac {n}{2^{e_2(n)}} + 1}{2}$. Then this one works: $ f(n) = \{\begin{array}{cc} t \mid m_n & (2(m_n - t) + 1)2^{e_2(n) + 1} \\ t \nmid m_n & (2m_n + 1)2^{e_2(n)} \\ \end{array}$