Solution 1. Let x be the number of times colors A and B meet, y the number of times colors A and C meet, and z the number of times colors B and C meet. If we end up with all of color A, then we have the system of three linear equations 13 - x - y + 2z = 45, 15 - x + 2y - z = 0, 17 + x - y - z = 0. This system has no solution all in integers, for example by Gaussian elimination. The same reasoning works for right-hand sides 0, 45, 0 and for 0, 0, 45. So it is impossible.

Solution 2. Let a, b, c be the number of chameleons of each color. Note that under any of the three possible changes, the value b - a changes by a multiple of 3. [a and b both decrease by 1 so b - a remains unchanged; or a decreases by 1 and b increases by 2 so b - a increases by 3; or a increases by 2 and b decreases by 1 so b - a decreases by 3.] But b - a begins at 2, so it can never reach 0, it can never reach 45, and it can never reach -45.

Solution 3. Represent the colors by 0, 1, and 2, for, respectively, the 13, 15, and 17 chameleons. Let a and b be the distinct colors of two chameleons who meet and let c be the third color. Then a + b = -c (mod 3). When the two chameleons change color, they both become color c, and c + c = 2c = -c (mod 3); so color changes do not change, modulo 3, the sum of the colors of all chameleons. Since the total number of chameleons is a multiple of 3, if all acquired the same color, then the color sum would be 0 (mod 3). However, the initial color sum is 13×0 + 15×1 + 17×2 = 1 (mod 3).