6. First construct a coloring method that does not meet the problem's requirements for $n \leqslant 16$.

Let $A_{1}, A_{2}, \cdots, A_{n}$ represent the vertices of a regular $n$-sided polygon (in clockwise order), and $M_{1}, M_{2}, M_{3}$ represent the sets of vertices of three different colors.

When $n=16$, let $M_{1}=\left\{A_{5}, A_{8}, A_{13}, A_{14}, A_{16}\right\}, M_{2}=\left\{A_{3}, A_{6}, A_{7}, A_{11}, A_{15}\right\}, M_{3}=\left\{A_{1}, A_{2}, A_{4}, A_{9}, A_{10}, A_{12}\right\}$. For $M_{1}$, the distance from $A_{14}$ to the other 4 vertices is different, and these 4 vertices form a rectangle. Similarly to $M_{1}$, it can be verified that $M_{2}$ does not contain 4 vertices that form the vertices of an isosceles trapezoid. For $M_{3}$, the 6 vertices are the endpoints of 3 diameters, so any 4 vertices either form the 4 vertices of a rectangle or the 4 vertices of a non-equilateral quadrilateral.

When $n=15$, let $M_{1}=\left\{A_{1}, A_{2}, A_{3}, A_{5}, A_{8}\right\}, M_{2}=\left\{A_{6}, A_{9}, A_{13}, A_{14}, A_{15}\right\}, M_{3}=\left\{A_{4}, A_{7}, A_{10}, A_{11}, A_{12}\right\}$, and no 4 points in each $M_{i}$ form the vertices of an isosceles trapezoid.

When $n=14$, let $M_{1}=\left\{A_{1}, A_{3}, A_{8}, A_{10}, A_{14}\right\}, M_{2}=\left\{A_{4}, A_{6}, A_{7}, A_{11}, A_{12}\right\}, M_{3}=\left\{A_{2}, A_{6}, A_{9}, A_{13}\right\}$, and no 4 points in each $M_{i}$ form the vertices of an isosceles trapezoid.

When $n=13$, let $M_{1}=\left\{A_{5}, A_{6}, A_{7}, A_{10}\right\}, M_{2}=\left\{A_{1}, A_{8}, A_{11}, A_{12}\right\}, M_{3}=\left\{A_{2}, A_{3}, A_{4}, A_{9}, A_{13}\right\}$, and no 4 points in each $M_{i}$ form the vertices of an isosceles trapezoid.

In the above cases, removing vertex $A_{13}$ and keeping the coloring method unchanged gives a coloring method for $n=12$; then removing vertex $A_{12}$ gives a coloring method for $n=11$; and continuing to remove vertex $A_{11}$ gives a coloring method for $n=10$.

When $n \leqslant 9$, the number of vertices of each color can be less than 4, so there are no 4 vertices of the same color that form the vertices of an isosceles trapezoid.

Therefore, $n \leqslant 16$ does not have the property required by the problem.
Next, we prove that $n=17$ satisfies the conclusion.
Proof by contradiction. Assume there exists a way to color the vertices of a regular 17-sided polygon with three colors such that no 4 vertices of the same color form the vertices of an isosceles trapezoid.

Since $\left[\frac{17-1}{3}\right]+1=6$, there must be 6 vertices of the same color, say yellow. Connecting these 6 points in pairs, we get $\mathrm{C}_{6}^{2}=15$ line segments. Since these line segments can have only $\left[\frac{17}{2}\right]=8$ different lengths, one of the following two cases must occur:
(1) There are 3 line segments of the same length.
Note that 3 < 17, so it is impossible for these 3 line segments to have a common vertex. Therefore, there must be two line segments with no common vertices. The 4 vertices of these two line segments satisfy the problem's requirements, leading to a contradiction.
(2) There are 7 pairs of line segments of the same length.

By the assumption, each pair of line segments of the same length must have a common yellow vertex, otherwise we could find 4 yellow vertices that satisfy the problem's requirements. By the pigeonhole principle, there must be two pairs of line segments with the same common yellow vertex. The other 4 vertices of these 4 line segments must form the vertices of an isosceles trapezoid, leading to a contradiction. Therefore, $n=17$ satisfies the conclusion.

In summary, the smallest value of $n$ is 17.