Van der Waerden's Theorem ensures that there is a positive integer $N$ such that if the terms of the sequence $1,2,\ldots,N$ are $3$-colored, then there is some monochromatic $3$-term arithmetic progression. (In fact, it's known that $27$ is the smallest $N$ with this property; it's one of the few van der Waerden numbers known.) Thus $3$-coloring the vertices of any one of the infinitely many regular $N$-gons inscribed in the circle yields a monochromatic isosceles triangle.

Consider a regular $13$-gon. By Pigeon Hole, there are $\left \lceil \frac {13}{3} \right \rceil = 5$ monochromatic corners. Let's number the corners as $1,2,\dots, 13$. If there is an arithmetic progression between any three corners in $\bmod 13$, these three corners will form an isosceles triangle. That is $z-x \equiv y - z \pmod {13}$, or $x+y \equiv 2z \pmod {13}$. If we prove the existence of isosceles triangle among $5$ monochromatic corners, we can get infinitely many polygons by rotating the regular $13$-gon with $\alpha \in \left [0,\tfrac {2\pi}{13}\right )$. Thus we can get infinitely many isosceles triangles. Let $P = \{x,y,z,u,v\}$ be the set of these monochromatic corners. Assume that none of three element subsets, say $a,b,c$, holds $a+b \equiv 2c \pmod {13}$. These $5$ numbers define $\binom {5}{2} = 10$ pairs: $x+y, x+z, x+u, x+v, y+z, y+u, y+v, z+u, z+v, u+v$. Some of these numbers can be equal. Let $S$ be the set of these numbers. None of the elements of $S$ belongs to $S' = \{2x,2y,2z,2u, 2v\}$. Since $2x \equiv 2y \Leftrightarrow x \equiv y \pmod {13}$ and $|P|=5$, number of elements of $S'$ is exactly $5$. Since $|S| + |S'| \leq 13$, we have $|S| \leq 8$. There are $10$ candidates for elements of $S$. So either there are $3$ distinct candidates such that $a \equiv b \equiv c \pmod {13}$ or there are $4$ distinct candidates such that $a \equiv b \ \land \ c \equiv d \pmod {13}$. First we prove the case $a \equiv b \equiv c \pmod {13}$ is impossible, then we prove the case $a \equiv b \ \land \ c \equiv d \pmod {13}$ conflicts with the above assumption. It cannot be: $x+y \equiv y +z \pmod {13} \Longrightarrow x \equiv z \pmod {13}$. Since $|P|=5$, we cannot find $3$ disjoint pairs. (At least $6$ elements are required for this.) For the second case: $a \equiv b \ \land \ c \equiv d \pmod {13}$ Let $a = x+y$, $b=z+u$. If $c = x+z$, either $d = y + u$ or $d = y+v$ or $d = u+v$. $d = y+ u$ does not hold; since $a+c \equiv b+d$ then $2x \equiv 2u \Rightarrow x \equiv y \pmod {13}$. For $d = y +v$, we get $a+c \equiv b+d$ then $2x \equiv u + v \pmod {13}$. Contradiction with the assumption. For $d = u+v$, we get $b+x \equiv a+d$ then $2z \equiv y + v \pmod {13}$. Contradiction with the assumption. So $S$ contains at least one element of $S'$. We can always find a monochromatic isosceles triangle among the corners of a regular $13$-gon.