I'll assume that k is a field and k(x) is a field of rational functions in the variable x. Unfortunately, , and do not form a group of automorphisms (it is not even closed), so you can't find the corresponding field. However, it seems like , and generate a group A of automorphisms (isomorphic to a symmetric group ), , where , , and .

So your base field is F=k(I) where . Now consider the splitting field for an irreducible polynomial f in k(I) such that , where (verify this). We see that .

Now we shall find an intermediate field corresponding a subgroup of . Let . It is easy to check that H is a subgroup of . The field corresponding H is , which is fixed by elements of H.

I'll leave it to you to find all other intermediate fields, because it seems like a lengthy computation is required to find and verify it.