Five. (20 points) Let $N=\{0,1,2, \cdots\}$, and given a $k \in \mathbf{N}$. Try to find all functions $f: \mathbf{N} \rightarrow \mathbf{N}$, such that for every $n \in \mathbf{N}$, we have
$$
f(n)<f(n+1) \text {, and } f(f(n))=n+2 k \text {. }
$$
Five, first prove the following lemma.
Lemma If the function $f: \mathbf{N} \rightarrow \mathbf{N}$ satisfies $f(n)<f(n+1)$ for every $n \in \mathbf{N}$, then for any $m, n \in \mathbf{N}, m \geqslant n$, we must have
$$
f(m)-f(n) \geqslant m-n \text {, and } f(m) \geqslant m.
$$

Proof of the lemma: Since $f(n)<f(n+1)$, and $f(n), f(n+1)$ are non-negative integers, we have
$$
\begin{array}{l}
f(n+1)-f(n) \geqslant 1, \\
f(n+2)-f(n+1) \geqslant 1, \\
\cdots \cdots \\
f(m)-f(m-1) \geqslant 1.
\end{array}
$$

Adding the above $m-n$ inequalities, we get
$$
f(m)-f(n) \geqslant m-n.
$$

Taking $n=0$ in the above inequality, we get
$$
f(m) \geqslant m+f(0) \geqslant m.
$$

Now, we prove the original problem.
From $f(f(n))=n+2k$ and the lemma, we have
$$
n+2k-f(n)=f(f(n))-f(n) \geqslant f(n)-n,
$$

i.e., $f(n) \leqslant n+k (n \in \mathbf{N})$.
In equation (1), replace $n$ with $f(n)$ to get
$$
n+2k=f(f(n)) \leqslant f(n)+k,
$$

i.e., $f(n) \geqslant n+k (n \in \mathbf{N})$.
From equations (1) and (2), we get $f(n)=n+k (n \in \mathbf{N})$. Clearly, this function satisfies the problem's requirements.
In summary, the function we seek is
$$
f(n)=n+k (n \in \mathbf{N}).
$$