= Solution
There is a separate problem with the printed linear-span assertion, when $Z_{((n-1),1)}$ denotes the usual <centralizer of a subalgebra>. Let $A_{n-1}$ be the sum of all <transpositions> in $H$. Direct expansion in the <group algebra> gives
$$
\Pi_n(X_n)=(n-1)1,\qquad
\Pi_n(X_n^2)=(n-1)1+2A_{n-1}.
$$
Thus $Y=X_n^2-2A_{n-1}$ commutes with $H$ and satisfies $\Pi_n(Y)=(n-1)1$. For $n\geq3$ it contains nonzero coefficients on three-cycles involving $n$, whereas every element of $\operatorname{span}\{1,X_n\}$ is supported only on the identity and <transpositions>. Therefore $Y$ is an explicit counterexample to the asserted equality.
One correct description of the entire preimage is obtained by putting $C=C_{\mathbb C S_n}(\mathbb C H)$. Equivariance gives $\Pi_n(C)\subseteq Z(\mathbb C H)$, and $\Pi_n$ is the identity on that center. It follows that
$$
\boxed{\Pi_n^{-1}(\mathbb C1)\cap C=\mathbb C1\oplus\ker(\Pi_n|_C).}
$$
A useful corrected two-dimensional statement restricts the support to the identity and the <transpositions> $(i\ n)$: invariance under conjugation by $H$ then forces their coefficients to be equal, yielding exactly $\operatorname{span}\{1,X_n\}$. This additional support restriction is absent from the PDF.
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