= Solution
Write $X=\begin{pmatrix}A&B\\C&D\end{pmatrix}$. Commuting with all $\operatorname{diag}(U_1,U_2)$ requires
$$
AU_1=U_1A,\quad DU_2=U_2D,\quad BU_2=U_1B,\quad CU_1=U_2C.
$$
For $n_1,n_2\geq2$, the defining <complex representations> of each factor are <irreducible>. <Schur lemma> gives $A=zI_{n_1}$ and $D=wI_{n_2}$. The two block <vector spaces> are inequivalent <representations> of the product <group>, even when $n_1=n_2$: the first factor acts nontrivially on just the first block. Thus their off-diagonal <intertwiners> vanish, $B=C=0$. <Unitarity> and the unit <determinant> give
$$
\boxed{C_G(H)=\{\operatorname{diag}(zI_{n_1},wI_{n_2}):|z|=|w|=1,\ z^{n_1}w^{n_2}=1\}.}
$$
\b[The assertion that this full group is always a circle needs correction.] Put $d=\gcd(n_1,n_2)$, $a=n_1/d$, $b=n_2/d$. The continuous <group homomorphism>
$$
(z,w)\longmapsto z^aw^b
$$
maps this <centralizer> onto the $d$th <roots of unity>. Its <kernel> is the <identity component of a Lie group>, parametrized injectively by $e^{i\theta}\mapsto(e^{ib\theta},e^{-ia\theta})$. Coprimality of $a,b$ proves injectivity; a <Bezout identity> proves that this parametrization exhausts the <kernel>. Extending the primitive integer row $(a,b)$ to a determinant-one integer matrix gives coordinates in which the constraint is simply $t^d=1$. Consequently
$$
\boxed{C_G(H)\cong U(1)\times\mathbb Z_d,\qquad C_G(H)^0\cong U(1).}
$$
For a concrete counterexample, take $n_1=n_2=2$. The constraint is $(zw)^2=1$, giving two separate circles $zw=1$ and $zw=-1$. The element $\operatorname{diag}(I_2,-I_2)$ lies in the second component, so this <centralizer> cannot be isomorphic as a <Lie group> to the connected <circle group>.
For the intended specialization of this <centralizer of two special unitary blocks>, $d=\gcd(2,3)=1$ and therefore the full <centralizer> really is a <circle group>:
$$
\boxed{U_c(1)=\{e^{i\theta X_c}:\theta\in\mathbb R/2\pi\mathbb Z\},\qquad
X_c=\operatorname{diag}(3,3,-2,-2,-2).}
$$
We use this primitive integer <U(1) charge> normalization below. If exactly one $n_i$ equals $1$ and the other is at least $2$, the same block-scalar formula still holds. If $n_1=n_2=1$, both factors are trivial and the <centralizer> is all of <SU(2)>, another exception to the printed claim.
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