Let $ H $ be a (closed) irreducible subgroup of $ SU(n) $. Suppose that $ K_1,K_2 $ are isomorphic subgroups of $ SU(n) $ that both contain $ H $. Must it be the case that $ K_1=K_2 $?
This question was inspired by observing in some small examples that if $ K_1 \cong N_{SU(n)}(H) \cong K_2 $ for $ H $ irreducible then $ K_1=K_2 $.
Note: this is not true for $ U(n) $ since there is an irreducible $ A_4 $ subgroup of $ U(3) $ (indeed it is a subgroup of $ SO(3) $) which can be extended to two different, even non-conjugate, $ S_4 $ subgroups. One is an $ S_4 $ subgroup of $ SO(3) $ (known as the octahedral group) while the other is an $ S_4 $ subgroup of $ O(3) $ (known as the full tetrahedral group)