 Solution

Write $D(z)=\mathrm{diag}(z^{-2}{I}_3,z^3{I}_2)$ and partition a matrix $X$ into $3+2$ blocks. The equation $XD(z)=D(z)X$ makes its off-diagonal blocks obey $(z^3-z^{-2})X_{12}=0$ and $(z^{-2}-z^3)X_{21}=0$ for every $z\in U(1)$. Choosing any $z$ with $z^5\ne 1$ forces both blocks to vanish. Conversely, every block diagonal matrix commutes with every $D(z)$. Intersecting with the special unitary group gives the exact centralizerIts Lie algebra is $\mathfrak{su}(3)\oplus \mathfrak{su}(2)\oplus \mathfrak{u}(1)$. At the group level there is a finite central identification. The surjective group homomorphismhas kernel $\{(z^2{I}_3,z^{-3}{I}_2,z):z^6=1\}\cong {\mathbb{Z}}_6$. Surjectivity follows by choosing $z$ with $z^{-6}=\det A^{\prime }$ for any target pair $(A^{\prime },B^{\prime })$, and setting $A=z^2{A}^{\prime }$, $B=z^{-3}{B}^{\prime }$. ThusThe six distinct kernel elements give a concrete reason the displayed map does not identify the centralizer with an unquotiented product.