Five, let $A=\{1,2,3, \cdots, 17\}$. For any function $f: A \rightarrow A$, denote
$$
f^{[1]}(x)=f(x), f^{[k+1]}(x)=f\left(f^{[k]}(x)\right)
$$
$(k \in \mathbb{N})$. Find the natural number $M$ such that:
$$
\begin{array}{l}
\text { (1) When } m<M, 1 \leqslant i \leqslant 16, \text { we have } \\
f^{[m]}(i+1)-f^{[m]}(i) \neq \pm 1(\bmod 17), \\
f^{[m]}(1)-f^{[m]}(17) \neq \pm 1(\bmod 17) ;
\end{array}
$$
(2) When $1 \leqslant i \leqslant 16$, we have
$$
\begin{array}{l}
f^{[M]}(i+1)-f^{[M]}(i) \\
\quad \equiv 1 \text { or }-1(\bmod 17), \\
\quad f^{[M]}(1)-f^{[M]}(17) \equiv 1 \text { or }-1(\bmod 17) .
\end{array}
$$

Determine the maximum possible value of $M$ for all functions $f$ that satisfy the above conditions, and prove your conclusion.
Five, the required $M_{0}=8$.
First, prove $M_{0} \geqslant 8$.
In fact, we can define the mapping $f(i) \equiv 3 i-2(\bmod 17)$, where $i \in A, f(i) \in A$.
If $f(i) \equiv f(j)(\bmod 17)$,
then $3 i-2 \equiv 3 j-2(\bmod 17)$,
we have $i \equiv j(\bmod 17)$,
$$
\therefore i=j \text {. }
$$

The mapping $f$ is a mapping from $A$ to $A$.
From the definition of the mapping $f$, we know
$$
\begin{array}{l}
f^{[n]}(i) \equiv 3^{n} \cdot i-3^{n}+1(\bmod 17) \text {. } \\
\text { If }\left\{\begin{array}{l}
f^{[M]}(i+1)-f^{[M]}(i) \equiv 1 \text { or }-1(\bmod 17), \\
f^{[M]}(1)-f^{[M]}(17) \equiv 1 \text { or }-1(\bmod 17),
\end{array}\right. \\
\text { then }\left\{\begin{array}{l}
{\left[3^{M}(i+1)-3^{M}+1\right]-\left[3^{M} \cdot i-3^{M}+1\right]} \\
\equiv+1 \text { or }-1(\bmod 17), \\
1 \quad\left[3^{M} \times 17-3^{M}+1\right]=+1 \text { or }-1(\bmod 17) .
\end{array}\right. \\
\end{array}
$$

Then $3^{M}=+1$ or $-1(\bmod 17)$
$$
\begin{aligned}
\text { Since } 3^{1} \equiv 3,3^{2} \equiv 9,3^{3} \equiv 10,3^{4}=13,3^{5}=5,3^{6} \equiv 15, \\
\left.3^{7} \equiv 11,3^{4} \equiv-1 \text { (mod } 17\right),
\end{aligned}
$$

Thus $M_{0} \geqslant 8$.
Proof: $M_{0} \leqslant 8$.
Consider a 17-sided polygon $A_{1} A_{2} \cdots A_{17}$, denoted as $G$. We first define:

When $i+1=18$, take $i+1$ as 1; when $i-1=0$, take $i-1$ as 17. Then, according to the rule, connect the segments: if $i \leqslant mp>q>0$, such that
$$
\begin{array}{l}
f_{[f]}^{[f]}(i)=f^{[\varphi]}(j), \\
f_{[f]}^{[f]}(i+1)=f^{[q]}(j+1) \\
f^{[p]}(i+1)=f^{[q]}(j-1) .
\end{array}
$$

Thus $f^{[\rho]}(i+1)=f^{[q]}(j-1)$.
From this, we have
$$
\left\{\begin{array}{l}
f^{[p-q]}(i)=j, \\
f^{[p-q]}(i+1)=j+1 \text { or } f^{[p-q]}(i+1)=j-1,
\end{array}\right.
$$

Therefore, the connected segments do not overlap.
Since $G$ has $17 \times 7$ diagonals.
Thus, $17 \times\left(M_{0}, 1\right) \leqslant 17 \times 7$.
Thus $M=8$.