Here are some conjectures and theorems that I and my group found:
For the 2 chameleons problem, if the opposite color of the color you want to make all of has an even number of chameleons, then it is possible. Otherwise it is not possible. (Theorem)
If there are an even number, then you can switch and get all of them close and then flip all of them 2 at a time. Otherwise, if you take 2 different then the numbers don't change. But if you do, then the number will stay odd. But 0 is even.
Assume a as a red and b as a blue.
aba to bbb:
aba, aab, bbb
For the 3 chameleons problem, assuming there are a red, b blue and c yellow with a>=b>=c, if a+b+c is divisible by 3 and a-b is divisible by 3 then all is possible. If a+b+c is divisible by 3 and a-b is not divisible by 3 then none is possible. Else 1 is possible. (Conjecture)
Assume there are a red, b blue and c yellow.
a=1, b=1, c=1: All in one step.
a=2, b=1, c=0: Only can go to a repeat.
a=2, b=1, c=1: a=4, b=0, c=0 but no others (check all cases).