Consider the specific element of the corresponding Lie algebra $\mathbb{1}\_3 \times \sigma^3$, where $\mathbb{1}\_3$ is the unit matrix in 3 dimensions, $\sigma^3$ is the 3rd Pauli matrix and $\times$ stands for Kronecker product. I want to find the space
$$
\left\{ U\in U(6) \; \text{ such that} \; U\left( \mathbb{1}\_3 \times \sigma^3\right)U^\dagger = \mathbb{1}\_3 \times \sigma^3 \right\}.
$$
I am particularly interested in whether this set is connected or disjoint.



Set of $U(6)$ elements which leave a Lie-algebra element invariant under conjugation
It is a theorem (of Hopf, I believe) that the centralizer of any member of the Lie algebra (not just $\mathbb{1}\_3 \otimes \sigma^3$) is connected. See Bourbaki, Lie groups, Chap. 9, §2, nº2, [Corollary 5](https://books.google.com/books?id=oCO0xOzNLhAC&pg=PA290).


In your case $\mathbb{1}\_3 \otimes \sigma^3$ is conjugate to $\smash[b]{\begin{pmatrix}\mathbb{1}\_3&0\\0&-\mathbb{1}\_3\end{pmatrix}}$ and so the centralizer is conjugate to the diagonal $U(3)\times U(3)$.


