[Solution] Such a function exists.
We take $\alpha=\frac{\sqrt{5}+1}{2}, \beta=\frac{\sqrt{5}-1}{2}$, clearly $\alpha=1+\beta, \alpha \beta=1$, let $f(n)=[\alpha n+\beta], n \in N$. We prove that $f(n)$ satisfies the conditions $f: N \rightarrow N$, and
$$\begin{array}{l} f(1)=[\alpha+\beta]=[\sqrt{5}]=2, \\ \begin{aligned} f(n+1) & =[(n+1) \alpha+\beta] \\ & =[\alpha n+\beta+\alpha] \geqslant[\alpha n+\beta+1] \\ & =f(n)+1>f(n) . \\ f(f(n)) & =[\alpha[\alpha n+\beta]+\beta] \\ & =[\alpha n+\beta]+[\beta[\alpha n+\beta]+\beta] \\ & =f(n)+[\beta[\alpha n+\beta]+\beta] . \end{aligned} \end{array}$$

Clearly, $\alpha n+\beta$ is not an integer, thus,
$$\begin{aligned} & \beta[\alpha n+\beta]+\beta & \beta(\alpha n+\beta-1)+\beta \\ = & n+\beta^{2}>n . \end{aligned}$$

So $[\beta[\alpha n+\beta]+\beta]=n$.
From (1) we get
$$f(f(n))=f(n)+n$$

Therefore, $f(n)=[\alpha n+\beta]$ satisfies all the conditions.
Note The function $f(n)$ sought is not unique, there are multiple construction forms, such as:
(A) $f(n)=\left[\frac{\sqrt{5}+1}{2} n+\frac{1}{2}\right]$;
(B)
$$\begin{array}{l} f(1)=2 \\ f(n)=n+\max \{i<n: f(i) \leqslant n\} \end{array}$$
The functions constructed by (A) and (B) can both be verified to satisfy the conditions given in the problem.