Suppose that , this means we can think of as . Suppose that is an automorphism of . The mappingis completely determinedby (be sure you understand why this is). Now can take possible values: . It cannot be because since is an automophism it maps . This means there are possible values for : . Let be the mapping determined by , then it means , i.e. is the identity mapping. Let be the mapping determined by , then it means . Let be the mapping determined by , then it means . Finally let be . These are all automorphism of and they form a group. Note that this group has an element of order , also note that while and . This means that is a cyclic group generated by (or ). Thus, we can think of this automorphism group as . Now to complete this problem we can think of as (where ) is its generator. If is a homomorphism thenit is completely determinedby thus , it cannot be (the identity) thus there are just three possibilities . Do the similar computation done above to get your answer.