Suppose there are $M$ different colors of ball and you have $N$ of each kind. You arrange all of the balls in an arbitrary linear order. Is it always possible to find a permutation of order 2 such that no ball shares a color with its neighbor?
(Note: a permutation of order 2 is a permutation which is its own inverse.)
If you only have one color $(M=1)$, then excluding the trivial one-ball case, it's impossible. If you have two colors $(M=2)$, it's always possible: supposing the colors are black and white, just swap every black ball in an odd-numbered position with a white ball in an even-numbered position.
The case for $M>2$ colors seems tricky to me, because the available moves (permutation) and success condition (no same-color neighbors) seem incomparable. I have tried extending the above $M=2$ strategy to greater numbers of colors without much success (e.g. considering index mod $M$, or trying to interleave pairs of colors.) I have also tried to represent the list of colored balls as integers mod $M$, or even the differences between neighboring balls as integers mod $M$; then, swaps affect the difference list in a particular way.
I have tried representing the list of balls as a matrix $A$, where $A_{i,j}$ is the number of balls of color $i$ in a position that is equal to $j\pmod{M}$; then, the sums of the rows and columns are all $N$, and one special way of achieving the desired condition is if the colors are all interleaved so that the matrix becomes a multiple of the identity matrix.