There are exactly 3 nontrivial proper normal subgroups of $ SO_4 $. This can be seen by using Goursat's lemma on the double cover $ SU_2 \times SU_2 $ and then noting that only 5 of the 10 normal subgroups in $ SU_2 \times SU_2 $ descend to the quotient $ SO_4 $ (in other words only 5 contain the element $ (-1,-1) $ ). Of these five normal subgroups, two are just the trivial group and all of $ SO_4 $. The remaining groups are the center $ \pm I $ and the groups of left and right isoclinic rotations, respectively, both of which are isomorphic to $ SU_2 $. Are these two copies of $ SU_2 $ characteristic subgroups? (If not then the outer automorphism given by conjugation by a reflection in $ O_4 $ must switch the two copies of $ SU_2 $.)
Also, consider the following. The complex numbers $ \mathbb{C} $ are a 2d real vector space. $ SO_2 $ acts on $ \mathbb{C} $ by rotations. There is a homomorphism $ SO_2 \to U_1 $ given by $$ \phi \mapsto \phi(1) $$ The quaternions $ \mathbb{H} $ are a 4d real vector space. $ SO_4 $ acts on $ \mathbb{H} $ by rotations. There is a map $ R: SO_4 \to SU_2 $ (here $ SU_2 $ is thought of as the group of norm 1 quaternions) given by $$ R: \phi \mapsto \phi(1) $$ This map takes the identity to the identity. Is it a group homomorphism? If so is the kernel would have to be $ SU_2 $ by my argument above. But that doesn't seem right. Is there any easy way to see this is not a homomorphism? What nice properties does this map have?
Question: Are the groups of left and right isoclinic rotations characteristic? Or can they be exchanged by the outer automorphism of $ SO_4 $? Is one of these groups the kernel of the map $ R $? Is $ R $ even a homomorphism?