Let $f$ be the solution. Then
$$f(f(n)) \leq \max\{n, f(n)\}, \forall n \in \mathbb{N}. \quad (*)$$
Let $f^k(a) = \underbrace{f(f(...f(a)))}_{k}$. Now there is $a$ such that $a > f(a)$. Then $f^2(a) < a$; by induction to $k$ it's easy to prove that
$$f^k(a) < a, \forall k > 0. \quad (1)$$
Because the set $\{1, 2, ..., a-1\}$ is finite, there is $i, j$ such that $0 < i < j$ and
$$f'(a) = f'(a) = f'(f^{j-i}(a)).$$
The function $f$ is injection so $f^{j-i}(a) = a$, which is contradiction to (1). Then
$$a \leq f(a), \forall a > 0 \quad (2)$$
and $f(a) \leq f(f(a)), \forall a \in \mathbb{N}$.
On another side, from (*) and (2) we have $f(f(a)) \leq \max\{a, f(a)\} = f(a)$ i.e. $f(f(a)) = f(a)$.
Finally $f(a) = a, \forall a \in \mathbb{N}$. So $f(n) = n, \forall n \in \mathbb{N}$.