Let $R$, $G$, $B$, and $Y$ denote the sets of Red, Green, Blue, and Yellow points, respectively, and let $r$, $g$, $b$, and $y$ denote a generic Red, Green, Blue, and Yellow point, respectively. For $0 \le k \le 431$, denote by $\mathcal{T}_k$ the counterclockwise rotation by $\frac{360k}{432}$ degrees around the center of the circle.

First, we claim that there is some index $i_1$ such that $|\mathcal{T}_{i_1}(R) \cap G| \ge 28$. Indeed, for each $k$, the set $\mathcal{T}_k(R) \cap G$ consists of all Green points that are the images of Red points under the rotation $\mathcal{T}_k$. Hence the sum
$$s_1 = |\mathcal{T}_0(R) \cap G| + |\mathcal{T}_1(R) \cap G| + \dots + |\mathcal{T}_{431}(R) \cap G|$$
is equal to the number of pairs of points $(r, g)$ such that $g = \mathcal{T}_k(r)$ for some $k$. On the other hand, for each $r$ and each $g$, there is a unique rotation $\mathcal{T}_k$ with $\mathcal{T}_k(r) = g$, from which it follows that $s_1 = 108^2 = 11664$. On the other hand, note that $|\mathcal{T}_0(R) \cap G| = |R \cap G| = 0$ because the sets $R$ and $G$ are disjoint. By the Pigeonhole principle, there is some index $i_1$ such that
$$|\mathcal{T}_{i_1}(R) \cap G| \ge \left\lfloor \frac{s_1}{431} \right\rfloor = \left\lfloor \frac{11664}{431} \right\rfloor = \lceil 27.06 \dots \rceil = 28,$$
establishing the claim. Let $RG$ denote the set $\mathcal{T}_{i_1}(R) \cap G$, and let $rg$ denote a generic point in $RG$.

Second, we claim that there is some index $i_2$ such that $|\mathcal{T}_{i_2}(RG) \cap B| \ge 8$. Again, for each $k$, the set $\mathcal{T}_k(RG) \cap B$ consists of all Blue points that are the images of the points in $RG$ under the rotation $\mathcal{T}_k$. Hence the sum
$$s_2 = |\mathcal{T}_0(RG) \cap B| + |\mathcal{T}_1(RG) \cap B| + \dots + |\mathcal{T}_{431}(RG) \cap B|$$
is equal to the number of pairs of points $(rg, b)$ such that $b = \mathcal{T}_k(rg)$ for some $k$. On the other hand, for each $rg$ and each $b$, there is a unique rotation $\mathcal{T}_k$ with $\mathcal{T}_k(rg) = b$, from which it follows that $s_2 \ge 28 \cdot 108 = 3024$. Clearly, $RG$ is a subset of $G$, which is disjoint from $B$, so $\mathcal{T}_0(RG) \cap B = \emptyset$. Furthermore, $\mathcal{T}_{432-i_1}(\mathcal{T}_{i_1}(R)) = R$. Therefore, because $\mathcal{T}_{432-i_1}(RG)$ is a subset of $R$, it is also disjoint from $B$, so $\mathcal{T}_{432-i_1}(RG) \cap B = \emptyset$. By the Pigeonhole principle, there is some index $i_2$ such that
$$|\mathcal{T}_{i_2}(RG) \cap B| \ge \lfloor \frac{s_2}{430} \rfloor \ge \lfloor \frac{3024}{430} \rfloor = \lceil 7.0325 \dots \rceil = 8,$$
establishing the claim. Let $RGB$ denote the set $\mathcal{T}_{i_2}(RG) \cap B$, and let $rgb$ denote a generic point in $RGB$.

Finally, we claim that there is some index $i_3$ such that $|\mathcal{T}_{i_3}(RGB) \cap Y| \ge 3$. We repeat the machinery from our previous arguments one more time to show that
$$s_3 = |\mathcal{T}_0(RGB) \cap Y| + |\mathcal{T}_1(RGB) \cap Y| + \dots + |\mathcal{T}_{431}(RGB) \cap Y| \ge 8 \cdot 108 = 864$$
and
$$|\mathcal{T}_0(RGB) \cap Y| = |\mathcal{T}_{432-i_2}(RGB) \cap Y| = |\mathcal{T}_{432-i_2-i_1}(RGB) \cap Y| = 0.$$
The Pigeonhole principle then shows that there is some index $i_3$ such that
$$|\mathcal{T}_{i_3}(RGB) \cap Y| \ge \lfloor \frac{s_3}{429} \rfloor \ge \lfloor \frac{864}{429} \rfloor = \lceil 2.01 \dots \rceil = 3,$$
establishing the claim.

We are now ready to construct the desired configuration. Let $y_1$, $y_2$, $y_3$ be three distinct points in $\mathcal{T}_{i_3}(RGB) \cap Y$. By the definition of sets $R$, $RG$, and $RGB$, the triples of points
$(y_1, y_2, y_3), \quad \mathcal{T}_{432-i_3}(y_1, y_2, y_3), \quad \mathcal{T}_{432-i_3-i_2}(y_1, y_2, y_3), \quad \text{and} \quad \mathcal{T}_{432-i_3-i_2-i_1}(y_1, y_2, y_3)$
form congruent triangles whose vertices are Yellow, Blue, Green, and Red, respectively, yielding the desired configuration of monochromatic triangles.