I start you out with and you think about . Since is the splitting of over it means . We will prove that . If is an automorphism of then it is completely determined by and by . Now is a zero of the irreducible polynomial , this polynomial has zeros , thus . By similar reasoning since is a zero of the irreducible polynomial . By isomorphism extension theorem there is an automorphism so that and . Also, there is an automorphism so that and . Notice that so is in fact complex-conjugation. We see that the order of in the Galois group is and order of in the Galois group is , furthermore . You want, as the problem says, to find all so that . Remember by the fundamental theorem of Galois theory where is the corresponding subgroup i.e. . Thus, corresponds to the subgroup of which has index i.e. that is and there is only one such subgroup. Therefore, there is only one subfield of which has . It is clear to see that and so .