= Solution
Write $D(z)=\operatorname{diag}(z^{-2}I_3,z^3I_2)$ 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 $z^5\ne1$ forces both blocks to vanish. Conversely, every block diagonal <matrix> commutes with every $D(z)$. Intersecting with the <special unitary group> gives the exact <centralizer>
$$
\boxed{C_{SU(5)}(U(1))=S(U(3)\times U(2))
=\{\operatorname{diag}(A,B):A\in U(3),\ B\in U(2),\ \det A\det B=1\}.}
$$
Its <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 homomorphism>
$$
F:SU(3)\times SU(2)\times U(1)\longrightarrow S(U(3)\times U(2)),\qquad
F(A,B,z)=\operatorname{diag}(z^{-2}A,z^3B)
$$
has <kernel> $\{(z^2I_3,z^{-3}I_2,z):z^6=1\}\cong\mathbb Z_6$. Surjectivity follows by choosing $z$ with $z^{-6}=\det A'$ for any target pair $(A',B')$, and setting $A=z^2A'$, $B=z^{-3}B'$. Thus
$$
\boxed{C_{SU(5)}(U(1))\cong
\bigl(SU(3)\times SU(2)\times U(1)\bigr)/\mathbb Z_6.}
$$
The six distinct <kernel> elements give a concrete reason the displayed map does not identify the <centralizer> with an unquotiented product.
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