Quote:
Originally Posted by Walter
I have some solutions, but it felt way too easy, so I am pretty sure I am missing something and would like to check my understanding of the problem:
First of all, for the RPS(5) game, is the "permutation" that changes none of the labels also counted as an automorphism or not? I only found 4 permutations that change at least one of the labels for the given example.
I am also confused by how the permutations are defined and what permutations are actually allowed in this case. Let's assume I have a permutation that relabels the numbers as follows:
0 to 1
1 to 2
2 to 0
3 to 4
4 to 3
I can't define this permutation in a single list. If we draw this in a graph, we get a disconnected graph with two cycles. Are we limited to permutations that result in a single cycle?

Yes, the identity mapping is counted as a permutation.
And the mapping can have two disconnected cycles, but I doubt that it would yield a solution.
Here is how you should define the mapping: [1,2,0,4,3] where the indices are the original numbers and the entries are the result of the permutation. You just have to be careful in applying this permutation to the game: do not apply one by one otherwise you will mess things up. It has to be applied all at once.
Let's apply your mapping to the game given in the problem:
0 > 1, 3
1 > 2, 4
2 > 0, 3
3 > 1, 4
4 > 0, 2
Here is the result:
1 > 2, 4
2 > 0, 3
0 > 1, 4
4 > 2, 3
3 > 1, 0
when sorted by the left hand side this yields the game in canonical form:
0 > 1, 4
1 > 2, 4
2 > 0, 3
3 > 1, 0
4 > 2, 3
I hope that it is clear now.