Question: What is the centralizer and normalized of $U(n)$ in $SU(n+1)$?
(Thanks in advance for your comments/answers!)
We can embed $U(n)$ in $SU(n+1)$ in the following way:
Take $U(n)=\frac{SU(n) \times U(1)}{\mathbb{Z}/n}$, and write the group element in $U(n)$ as $g_{U(n)} \in U(n)$ such that $$ g_{U(n)} =(a_{SU(n)}, b_{U(1)} ) $$ with the constraint that $$ \det a_{SU(n)}=1. $$ The $\mathbb{Z}/n$ center of $SU(n)$ is identified with the subgroup $\mathbb{Z}/n \in U(1)$. For example, $$\text{ if } a_{SU(n)}= \exp(\frac{2 \pi i}{n})\mathbb{I}_{n\times n}, \text{ then } b_{U(1)} = \exp(\frac{2 \pi i}{n}) \text{ is identified with as the same group action by } a_{SU(n)}. $$ Now the $g_{U(n)} =(a_{SU(n)}, b_{U(1)} ) \in U(n)$ can be written as the group element $$ g_{SU(n+1)} = \begin{pmatrix} g_{U(n)} & 0\\ 0 & (b_{U(1)})^{-n} \end{pmatrix} \in SU(n+1) $$ so that $\det g_{SU(n+1)}=\det g_{U(n+1)}(b_{U(1)})^{-n}=b_{U(1)}^n b_{U(1)}^{-n}=1$ as desired.
So the embedding $U(n)\to SU(n+1)$ is simply $g\mapsto[\begin{smallmatrix}g&0\\0&(\det g)^{-1}\end{smallmatrix}]$.
Suppose $x\in N_{SU(n+1)}(U(n))$. For any $\tau\in U(n)$, we know $x\tau x^{-1}$ is in $U(n)$, so $x\tau x^{-1}$ must have $e_{n+1}$ as an eigenvector, which implies $x^{-1}e_{n+1}$ must be a simultaneous eigenvector of all $\tau\in U(n)$. That implies that $x^{-1}e_{n+1}$ is a multiple of $e_{n+1}$, or in other words $e_{n+1}$ is a simultaneous eigenvector of all $x$ in the normalizer, which in turn implies $N_{SU(n+1)}(U(n))=U(n)$, i.e. $U(n)$ is self-normalizing.
Thus the centralizer is the center $Z(U(n))$. This consists of $[\begin{smallmatrix}\lambda I_n &0\\0& \lambda^{-n} \end{smallmatrix}]$ with $\lambda\in S^1$.