Let $ \Gamma $ and $ \Gamma' $ be two octahedral subgroups (isomorphic to the 24 element group $ S_4 $) of $ SO_3(\mathbb{R}) $. Then any isomorphism $$ \phi: \Gamma \to \Gamma' $$ extends to a unique automorphism of $ G=SO_3(\mathbb{R}) $. To see this note that all octahedral subgroups of $ G $ are conjugate (indeed this is true for all isomorphic finite subgroups of $ G $). So there always exists some $ g \in G $ such that $ g \Gamma g^{-1}=\Gamma' $ giving us the correct image and so agreeing with $ \phi $ up to some automorphism. From there we can conjugate by some $ x \in \Gamma \cong Aut(\Gamma) $ to get an exact match. So that conjugation by $ xg $ is an automorphism of $ G $ extending $ \phi $.
From there we can get uniqueness by noting that every automorphism of of $ G $ is inner and that if $ g_1,g_2 $ both conjugate $\Gamma $ to $ \Gamma' $ through exactly the action of $ \phi $ then conjugation by $ g_1g_2^{-1} $ fixes $ \Gamma $ pointwise. So $ g_1g_2^{-1} \in N_G(\Gamma) $ but $ \Gamma $ is self-normalizing $ N_G(\Gamma)=\Gamma $ so $ g_1g_2^{-1} \in \Gamma $. However $ \Gamma $ also has trivial center $ Z(\Gamma)=1 $ so we must have $ g_1=g_2 $ and thus uniqueness is proven.
This leads me to ask:
Let $ G $ be a compact connected Lie group and $ \phi:S_4 \to G $ an injective group homomorphism. What can we say about extensions $ \tilde{\phi} : SO_3(\mathbb{R}) \to G $? Do they always exist? Are they always unique?
