Every open subgroup of the Nottingham group (p>3) is both Hopfian and co-Hopfian. On the one hand, the Nottingham group is hereditarily just infinite. So all it open subgroup are just inifinite, that is their proper quotients are finite. On the other hand, Mikhail Ershov showed that the commensurator of the Nottingham group (p>3) is its automorphism group. So if two open subgroups are isomorphic the isomorphism extends to an automorphism of the Nottingham group. In particular, the indices in the Nottingham group of such subgroups are the same. Thus, if one is a subgroup of the other, they are equal.

EDIT: I should be a bit more careful. The fact that the commensurator of the Nottingham group is its automorphism group, does not say that the isomorphism extends to an automorphism, but that the isomorphism restricted to an open subgroup extends to an automorphism. This is good enough for the claim above. It is also may be that Mikhail actually proved the stronger claim that I made (I am not 100% sure).

Every open subgroup of the Nottingham group (p>3) is both Hopfian and co-Hopfian. On the one hand, the Nottingham group is hereditarily just infinite. So all it open subgroup are just infinite, that is their proper quotients are finite. On the other hand, Mikhail Ershov showed that the commensurator of the Nottingham group (p>3) is its automorphism group. So if two open subgroups are isomorphic the isomorphism extends to an automorphism of the Nottingham group. In particular, the indices in the Nottingham group of such subgroups are the same. Thus, if one is a subgroup of the other, they are equal.

EDIT: I should be a bit more careful. The fact that the commensurator of the Nottingham group is its automorphism group, does not say that the isomorphism extends to an automorphism, but that the isomorphism restricted to an open subgroup extends to an automorphism. This is good enough for the claim above. It is also may be that Mikhail actually proved the stronger claim that I made (I am not 100% sure).

EDIT2: I have got confused about the definition of co-Hopfian. This argument shows that the Nottingham group is finite co-Hopfian. It is not true that it is co-Hopfian from results of Rachel Camina (and also a paper of Fesenko).

Every open subgroup of the Nottingham group (p>3) is both Hopfian and co-Hopfian. On the one hand, the Nottingham group is hereditarily just infinite. So all it open subgroup are just inifinite, that is their proper quotients are finite. On the other hand, Mikhail Ershov showed that the commensurator of the Nottingham group (p>3) is its automorphism group. So if two open subgroups are isomorphic the isomorphism extends to an automorphism of the Nottingham group. In particular, the indices in the Nottingham group of such subgroups are the same. Thus, if one is a subgroup of the other, they are equal.

Every open subgroup of the Nottingham group (p>3) is both Hopfian and co-Hopfian. On the one hand, the Nottingham group is hereditarily just infinite. So all it open subgroup are just inifinite, that is their proper quotients are finite. On the other hand, Mikhail Ershov showed that the commensurator of the Nottingham group (p>3) is its automorphism group. So if two open subgroups are isomorphic the isomorphism extends to an automorphism of the Nottingham group. In particular, the indices in the Nottingham group of such subgroups are the same. Thus, if one is a subgroup of the other, they are equal.

EDIT: I should be a bit more careful. The fact that the commensurator of the Nottingham group is its automorphism group, does not say that the isomorphism extends to an automorphism, but that the isomorphism restricted to an open subgroup extends to an automorphism. This is good enough for the claim above. It is also may be that Mikhail actually proved the stronger claim that I made (I am not 100% sure).