5. For $n \in \mathbf{N}$, there are two functions satisfying the conditions:
$f(n)=n+1$,
$$f(n)=\left\{\begin{array}{ll} n+1, & n \equiv 0 \text { or } 2(\bmod 4) ; \\ n+5, & n \equiv 1(\bmod 4) ; \\ n-3, & n \equiv 3(\bmod 4) . \end{array}\right.$$

Let $h^{0}(x)=x$,
$h^{k}(x)=\underbrace{h(\cdots h}_{k \uparrow}(x) \cdots))\left(k \in \mathbf{Z}_{+}\right)$.
From equation (1) we get
$f^{4}(n)=f\left(f^{3}(n)\right)=f(f(n+1)+1)$,
$f^{4}(n+1)=f^{3}(f(n+1))$
$=f(f(n+1)+1)+1$.
Thus, $f^{4}(n)+1=f^{4}(n+1)$.
(1) Let $R_{i}$ denote the range of $f^{i}$.
(ii) $b=f(0)+1$;
(iii) $b-1 \in S_{1}$.
Otherwise, $b-1 \in R_{1}$, and there exists $n \in \mathbf{Z}_{+}$ such that
$f(n)=b-1$.
Thus, $f^{3}(n-1)=f(n)+1=b$.
Therefore, $b \in R_{3}$, which is a contradiction.
By $3 k=\left|S_{1} \cup S_{2} \cup S_{3}\right| \leqslant 1+1+\left|S_{1}\right|=k+2$, we have
$k \leqslant 1$.
Thus, $k=1$, and the equality in the inequality holds.
Therefore, there exists some $a \in \mathbf{N}$ such that
$S_{1}=\{a\}, S_{2}=\{f(a)\}, S_{3}=\left\{f^{2}(a)\right\}$.

Since $f^{0}(x)=x$, then
the conditions (i), (ii), and (iii) each hold
$R_{0}=\mathbf{N}$, and $R_{0} \supseteq R_{1} \supseteq \cdots$.
From equation (3), if $a \in R_{4}$, then $a+1 \in R_{4}$.
This indicates that $\mathbf{N} \backslash R_{4}$ is finite, hence $\mathbf{N} \backslash R_{1}$ is finite. In particular, $R_{1}$ is unbounded.

If there exist different non-negative integers $m, n$ such that $f(m)=f(n)$, from equation (1) we get
$f(m+1)=f(n+1)$.
By mathematical induction, for every $c \in \mathbf{N}$, we have
$f(m+c)=f(n+c)$.
Thus, for all $k \geqslant m$, the function $f(k)$ is periodic with period $|m-n|$. Therefore, $R_{1}$ is bounded, which is a contradiction.
Thus, $f$ is injective.
(2) Let $S_{i}=R_{i-1} \backslash R_{i}$.
Then for all positive integers $i(i \leqslant 4)$, $S_{i}$ is finite.
On the other hand, since $f$ is injective, we have
$n \in S_{i} \Leftrightarrow f(n) \in S_{i+1}$.
By $f$ being bijective, $f$ is a bijection between $S_{i}$ and $S_{i+1}$,

Thus, $\left|S_{1}\right|=\left|S_{2}\right|=\cdots$, and we denote $\left|S_{i}\right|=k$.
If $0 \in R_{3}$, then there exists $n \in \mathbf{N}$ such that
$f(f(f(n)))=0$.
From equation (1) we get $f(n+1)=-1$, which is a contradiction.
Thus, $0 \in R_{0} \backslash R_{3}=S_{1} \cup S_{2} \cup S_{3}$, and $k \geqslant 1$.
For elements $b$ in $R_{0} \backslash R_{3}=S_{1} \cup S_{2} \cup S_{3}$, at least one of the following three conditions is satisfied:
(i) $b=0$;

Thus $f(0)=f^{3}(a)=f(a+1)+1=1$.
Therefore, $a=f(0)+1=2$.
From equation (5) we get $f(2)=3$.
Thus, $f(0)=1, f(2)=3, f(3)=0$.
We will prove by induction on $m$ that for all $n=4 k, 4 k+2,4 k+3(k \leqslant m)$ and for all $n=4 k+1$ $(k<m)$, equation (2) holds.
When $m=0$, the conclusion holds.
Assume the conclusion holds for $m-1(m \geqslant 1)$.
From equation (1) we get
$f^{3}(4 m-3)=f(4 m-2)+1=4 m$.
From equation (3) we get
$$\begin{array}{l} f(4 m)=f^{4}(4 m-3)=f^{4}(4 m-4)+1 \\ =f^{3}(4 m-3)+1=4 m+1 . \end{array}$$

By the induction hypothesis and equation (1) we get
$$\begin{array}{l} f(4 m-3)=f^{2}(4 m-4)=f^{3}(4 m-1) \\ =f(4 m)+1=4 m+2, \\ f(4 m+2)=f^{2}(4 m-3)=f^{3}(4 m-4) \\ =f(4 m-3)+1=4 m+3, \\ f(4 m+3)=f^{2}(4 m+2)=f^{3}(4 m-3) \\ =f(4 m-2)+1=4 m . \end{array}$$

In summary, equation (2) holds.
Direct verification shows that equation (2) is a solution to the original equation.
$$\begin{array}{l} f^{3}(4 k)=4 k+7=f(4 k+1)+1, \\ f^{3}(4 k+1)=4 k+4=f(4 k+2)+1, \\ f^{3}(4 k+2)=4 k+1=f(4 k+3)+1, \\ f^{3}(4 k+3)=4 k+6=f(4 k+4)+1 . \end{array}$$