Let $n$ and $m$ be two positive integers. There is a canonical inclusion $U(n) \rightarrow U(nm)$ given by the tensor product with the unit matrix $\mathbf{1}_m$.
What is the centralizer of $U(n)$ seen as a subgroup of $U(nm)$?
Edit: I was asked to provide context, so here it is, but I'm not sure it will be very illuminating for a non-physicist.
Context: In physical language, you can consider a field theory with $nm$ free complex scalar fields $\phi_i$, with $i=1,\dots,nm$. The Lagrangian density is then $\sum_i \partial_\mu \phi_i^* \partial^{\mu} \phi_i$. It is left invariant under the action of the unitary group $U(nm)$, the vector $(\phi_i)_{i=1 , \dots , nm}$ being in the fundamental representation.
Now you might want to gauge partially this big symmetry. Here this means you identify a subgroup $U(n) \subset U(nm)$ and consider the combinations of the $\phi_i$ that are invariant under $U(n)$. A way to visualize that is to see the fields $\phi_i$ as the entries of an $n \times m$ where $U(n)$ acts by left multiplication. Then it is obvious that there is also a group $U(m)$ which acts by right multiplication on this matrix, and the two actions commute. Therefore, the centralizer of $U(n)$ contains $U(m)$.
My question can then be re-formulated as: is the centralizer equal to $U(m)$, or bigger?
