The fewest possible moves in which this puzzle can be solved are 118. 

I will give the complete solution. 
The black figures on white discs move in the directions of the hands of a clock, and the white figures on black discs the other way. The following are the numbers in the order in which they move. 
Whether you have to make a simple move or a leaping move will be clear from the position, as you never can have an alternative.

The moves enclosed in brackets are to be played five times over: 6, 7, 8, 6, 5, 4, 7, 8, 9, 10, 6, 5, 4, 3, 2, 7, 8, 9, 10, 11 (6, 5, 4, 3, 2, 1), 6, 5, 4, 3, 2, 12, (7, 8, 9, 10, 11, 12), 7, 8, 9, 10, 11, 1, 6, 5, 4, 3, 2, 12, 7, 8, 9, 10, 11, 6, 5, 4, 3, 2, 8, 9, 10, 11, 4, 3, 2, 10, 11, 2. 
We thus have made 118 moves within the conditions, the black frogs have changed places with the white ones, and 1 and 12 are side by side in the positions stipulated.

The general solution in the case of this puzzle is 3n² + 2n - 2 moves, where the number of frogs of each colour is n. 
The law governing the sequence of moves is easily discovered by an examination of the simpler cases, where n = 2, 3, and 4.

If, instead of 11 and 12 changing places, the 6 and 7 must interchange, the expression is n² + 4n + 2 moves. If we give n the value 6, as in the example of the Frogs' Ring, the number of moves would be 62.