(Solution by A. Goloubitskaya.) Let $\Gamma$ be the circumcircle of the given regular hexagon. It is evident that $B$ is the center of $\Gamma$. Since all sides of the regular hexagon are equal, we have $2\alpha = \text{arc } A_1A_2 = \text{arc } A_2A_3 = \dots = \text{arc } A_9A_{10} = \text{arc } A_{10}A_1 = 36^\circ$. Since $A_1A_2 = A_9A_{10}$, $\angle A_{10}A_2A_1 = \angle A_1A_9A_{10} = \alpha$ and $\angle A_2A_1A_9 = \angle A_2A_{10}A_9 = 7\alpha$, it follows that the triangles $A_1A_2C$ and $A_{10}A_9C$ are equal, so $CA_1 = CA_{10}$. Hence the line $BC$ is the bisector of the segment $A_1A_{10}$. Then the line $BC$ contains the altitude, the median, and the bisectrix of the isosceles triangle $A_{10}BA_1$ ($BA_{10} = BA_1$). Hence, $\angle A_1BC = 0.5 \text{ arc } A_{10}A_1 = \alpha = 18^\circ$. Similarly, $\angle A_2BA = 0.5 \text{ arc } A_2BA_4 = 0.5 \text{ arc } A_2A_4 = 2\alpha = 36^\circ$. Since $\angle A_2BA_1 = 2\alpha = 72^\circ$, we have

$$\angle ABC = \angle ABA_2 + \angle A_2BA_1 + \angle A_1BC = 36^\circ + 36^\circ + 18^\circ = 90^\circ.$$

Since $\angle AA_2C = \angle A_5A_2A_{10} = 5\alpha = 90^\circ$, we have $\angle ABC + \angle AA_2C = 180^\circ$, so the points $A$, $B$, $C$, $A_2$ are concyclic. Then $\angle BCA = \angle BA_2A = 2\alpha = 36^\circ$ and $\angle CAB = 180^\circ - \angle ABC - \angle BCA = 54^\circ$.