$\DeclareMathOperator\SU{SU}$Consider the Lie group $\SU(6)$, its Lie algebra $\mathfrak{su}(6)$ and the $\mathfrak{su}(2)$ subalgebra spanned by $\mathbb{1}\_3 \otimes \sigma^i$, where $\sigma^i$ are the 3 Pauli matrices, $\mathbb{1}\_3$ is the unit matrix in 3 dimensions and $\otimes$ stands for Kronecker product. I want to find the space
$$
\left\{ U\in \SU(6) \; \text{ such that} \; U\left( \mathbb{1}\_3 o\times \sigma^3\right)U^\dagger \in \mathfrak{su}(2) \right\}.
$$
i.e. the set of $U$ such that $U\left( \mathbb{1}\_3 \otimes \sigma^3\right)U^\dagger = \mathbb{1}\_3 \otimes \sigma^i \alpha^i$, for $\alpha^i \in \mathbb{R}$.
I am mostly interested in whether this set is connected or disjoint.



Set of $\mathrm{SU}(6)$ matrices which conjugate $\mathbb{1}_3 \otimes \sigma^3$ subalgebra element into $\mathfrak{su}(2)$
The set in question is isomorphic to the fiber product $F\times^SH$ where $F=SU(2)$, $H=S(U(3)\times U(3))$ and $S=F\cap H=U(1)$. In particular, it is a connected manifold.


This can be seen as follows: Let $\xi=1\_3\otimes\sigma^3$ und $U\in\mathfrak{su}(6)$ with $U\xi U^{-1}\in\mathfrak{su}(2)$. Since $\xi$ and $U\xi U^{−1}$ are elements of $\mathfrak{su}(2)$ with the same norm there is an $A\in SU(2)$ such that $U\xi U^{-1}=A\xi A^{-1}$ (because $\mathrm{Ad}SU(2)=SO(3)$). Thus $B:=A^{-1}U$ is in the centralizer $H$ of $\xi$ in $\mathfrak{su}(6)$. So $U=AB\in FH$. Now $A$ and $B$ are unique up to an element of $F\cap H=S$ which proves the claim.


