Example 5 Find the smallest natural number $n$ such that: when any 5 vertices of a regular $n$-gon $S$ are colored red, there is always a line of symmetry $L$ of $S$ such that the reflection of each red point across $L$ is not a red point.
Let the regular $n$-sided polygon be $A_{1}, A_{2}, \cdots, A_{n}$, which has $n$ axes of symmetry.
Let the axis of symmetry passing through the midpoint of edge $A_{1} A_{n}$ be denoted as $L_{1}$, the axis of symmetry passing through $A_{1}$ as $L_{2}$, the axis of symmetry passing through the midpoint of edge $A_{1} A_{2}$ as $L_{3}$, the axis of symmetry passing through $A_{2}$ as $L_{4}, \cdots$, and the axis of symmetry passing through the midpoint of edge $A_{\left[\frac{n}{2}\right]} A_{\left[\frac{m+1}{2}\right]}$ as $L_{n}$ (where $A_{\left[\frac{n}{2}\right]}$ and $A_{\left[\frac{\text { 薮 }}{}\right]}$ may coincide). It is easy to see that $A_{i}$ and $A_{j}$ are symmetric with respect to $L_{m}$ if and only if $i+j=m(\bmod n)$ (with the convention that $A_{n+i}=A_{i}$).

For any set $X=\left\{a_{1}, a_{2}, \cdots, a_{t}\right\}$, define $X^{*}=\left\{a_{i}+a_{j} \mid 1 \leqslant i \leqslant j \leqslant t\right\}$. For $k \in\{3,4,5, \cdots, 13\}$, color 5 vertices of the regular $k$-sided polygon $A_{1}, A_{2}, A_{4}, A_{6}, A_{7}$ red (where the subscripts are understood modulo $k$, and if the number of distinct vertices is less than 5, color any number of points repeatedly to make a total of 5 red points). Let $P=\{1,2,4,6,7\}$, then $P^{*}=\{2,3,4, \cdots, 14\}$, so $P^{*}$ contains a complete residue system modulo $k$ (a set of $k$ consecutive natural numbers, understood modulo $k$), thus for any $m(1 \leqslant m \leqslant k)$, there must be some $x \equiv m(\bmod k)$ in $P^{*}$, i.e., there exist $i+j \equiv m(\bmod k)$, where $i, j \in P$. Therefore, $A_{i}$ and $A_{j}$ are two points symmetric with respect to $L_{m}$. So, for any $k \in\{3,4,5, \cdots, 13\}$, it does not meet the requirement, i.e., $n \geqslant 14$.

On the other hand, for a regular 14-sided polygon, color any 5 vertices $A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{5}}$ red, let $P=\left\{i_{1}, i_{2}, \cdots, i_{5}\right\}$, introduce a parameter: suppose $P$ contains $r$ odd numbers and $5-r$ even numbers, then the number of odd numbers in $P^{*}$ is: $r(5-r) \leqslant\left[\left(\frac{5}{2}\right)^{2}\right]=6$ (where the numbers in $P$ and $P^{*}$ are understood modulo 14, which does not change their parity). Thus, in $\{1,3,5,7,9,11,13\}$, there must be an odd number $m$ such that $m \notin P^{*}$, i.e., for any $i, j \in P, i+j \notin P^{*}, i+j \neq \equiv m(\bmod 14)$. Therefore, any two red points are not symmetric with respect to $L_{m}$. Hence, $n_{\min }=14$.