15. As shown in Figure 8, the line $O B$ is the graph of the linear function $y=2x$, and the coordinates of point $A$ are $(0,2)$. Find point $C$ on line $O B$ such that $\triangle A C O$ is an isosceles triangle. Determine the coordinates of point $C$.
Three, 15. As shown in Figure 11, if this isosceles triangle has $O A$ as one of its legs, and $A$ as the vertex, then
$$
A O=A C_{1}=2 \text {. }
$$

Let $C_{1}(x, 2 x)$, we get
$$
x^{2}+(2 x-2)^{2}=2^{2} \text {. }
$$

Solving for $x=\frac{8}{5}$.
Thus, $C_{1}\left(\frac{8}{5}, \frac{16}{5}\right)$.
If this isosceles triangle has $O A$ as one of its legs, and $O$ as the vertex, then
$$
O C_{2}=O C_{3}=O A=2 .
$$

Let $C_{2}\left(x^{\prime}, 2 x^{\prime}\right)$, we get $x^{\prime 2}+\left(2 x^{\prime}\right)^{2}=4$.
Solving for $x^{\prime}=\frac{2 \sqrt{5}}{5}$, we get $C_{2}\left(\frac{2 \sqrt{5}}{5}, \frac{4 \sqrt{5}}{5}\right)$.
Since point $C_{3}$ is symmetric to point $C_{2}$ about the origin, we get
$$
C_{3}\left(-\frac{2 \sqrt{5}}{5},-\frac{4 \sqrt{5}}{5}\right) \text {. }
$$

If the isosceles triangle has $O A$ as the base, then the y-coordinate of point $C_{4}$ is 1. Therefore, its x-coordinate is $\frac{1}{2}$, giving $C_{4}\left(\frac{1}{2}, 1\right)$.
Thus, there are 4 points $C$ that satisfy the conditions, with coordinates
$$
\begin{array}{l}
\left(\frac{8}{5}, \frac{16}{5}\right),\left(\frac{2 \sqrt{5}}{5}, \frac{4 \sqrt{5}}{5}\right), \\
\left(-\frac{2 \sqrt{5}}{5},-\frac{4 \sqrt{5}}{5}\right),\left(\frac{1}{2}, 1\right) .
\end{array}
$$