25
There is one obvious group action: the trivial action. Now, we consider a nontrivial action. Then, there exists an element of {1,2,3,4,5}{1,2,3,4,5}, say 11, not fixed by Z5Z5. However, the size of the orbit of 11would then be 55 (since the size of the orbit divides the order of Z5Z5 and it is not 11). Hence, this action is transitive as the orbit must be the whole {1,2,3,4,5}{1,2,3,4,5}. Therefore, this action is a permutation of {1,2,3,4,5}{1,2,3,4,5} on a circle (where 1∈Z51∈Z5 is rotation by 2π52π5). On the other hand, every permutation of {1,2,3,4,5}{1,2,3,4,5} on a circle can be made an action of Z5Z5 on this set. There are (5−1)!=24(5−1)!=24 such permutations. Hence, in total, there are 1+24=251+24=25 possible actions of Z5 on {1,2,3,4,5}
In how many ways can the group Z6Z6 act on {1,2,3,4,5,6}{1,2,3,4,5,6}?