The relative centralizer is generated by the previous centre and the last Jucys-Murphy element

By [F3] the element $X_n$ commutes with $\mathbb{C}[S_{n-1}]$; every $z\in Z(\mathbb{C}[S_{n-1}])$ commutes with every element of $\mathbb{C}[S_{n-1}]$ by definition of the centre. Hence $X_n$ and all of $Z(\mathbb{C}[S_{n-1}])$ lie in $Z(\mathbb{C}[S_n],\mathbb{C}[S_{n-1}])$, and since that set is a subalgebra, $⟨Z(\mathbb{C}[S_{n-1}]),X_n⟩\subseteq Z(\mathbb{C}[S_n],\mathbb{C}[S_{n-1}])$.

Write $E$ for the set of pairs $(\mu ,\lambda )$ with $\mu ⊢n-1$, $\lambda ⊢n$ and $\mu \subseteq \lambda$; equivalently $E=\{(\mu ,\mu +x):x\in \mathrm{Add}(\mu )\}$. For $\mu ⊢n-1$ let $W_{\mu }$ denote the $\mu$-isotypic component of the multiplicity-free sum $W:={⨁}_{\lambda ⊢n}{V}^{\lambda }$ restricted to $S_{n-1}$; by [F2] it is the direct sum of the copies $V^{\mu }\subseteq V^{\lambda }$ over the pairs $(\mu ,\lambda )\in E$.

Structure of the centralizer. By [F1] applied to $S_n$, an element $z\in \mathbb{C}[S_n]$ corresponds to a tuple $(z_{\lambda }{)}_{\lambda ⊢n}$ with $z_{\lambda }\in \mathrm{End}(V^{\lambda })$, and $z$ lies in $Z(\mathbb{C}[S_n],\mathbb{C}[S_{n-1}])$ exactly when each $z_{\lambda }$ commutes with the $S_{n-1}$-action on $V^{\lambda }$. By [F2] and [F1] the commutant of $S_{n-1}$ in $\mathrm{End}(V^{\lambda })$ is ${\prod }_{x\in \mathrm{Rem}(\lambda )}\mathbb{C}{\mathrm{i}\mathrm{d}}_{V^{\lambda -x}}$: restriction is multiplicity-free; any nonzero map between two irreducible summands would be an isomorphism (its kernel and image are invariant), contrary to their distinct shapes, so off-diagonal maps vanish; by Schur's lemma an $S_{n-1}$-endomorphism of each irreducible summand $V^{\lambda -x}$ is a scalar. Hence evaluation on the copies induces an algebra isomorphism $Z(\mathbb{C}[S_n],\mathbb{C}[S_{n-1}])\to {\mathbb{C}}^E$, $z\mapsto (z(\mu ,\lambda ){)}_{(\mu ,\lambda )\in E}$, where $z(\mu ,\lambda )$ is the scalar by which $z$ acts on the copy of $V^{\mu }$ inside $V^{\lambda }$; it is injective because zero scalars on every restriction summand make every operator $z_{\lambda }$ zero, and the Wedderburn product isomorphism in [F1] then gives $z=0$, and surjective by [F1] because the scalars on the finitely many copies may be prescribed independently.

The image $A$ of $⟨Z(\mathbb{C}[S_{n-1}]),X_n⟩$ under the isomorphism of step 2.1 is the unital subalgebra of ${\mathbb{C}}^E$ generated by the evaluations of the two generating pieces: an element $z\in Z(\mathbb{C}[S_{n-1}])$ acts on every copy $V^{\mu }$ by the central character value ${\omega }_{\mu }(z)$ of the irreducible $V^{\mu }$; and $X_n$ acts on the copy $(\mu ,\lambda )=(\mu ,\mu +x)$ by the scalar $c(x)$ by [F5].

Separation of edges with distinct predecessor. Let $(\mu ,\lambda ),({\mu }^{\mathrm{\prime }},{\lambda }^{\mathrm{\prime }})\in E$ with $\mu \mathrm{\ne }{\mu }^{\mathrm{\prime }}$. The central idempotent $e_{\mu }\in Z(\mathbb{C}[S_{n-1}])$ of the Wedderburn factor $\mathrm{End}(V^{\mu })$ acts as $1$ on $V^{\mu }$ and as $0$ on $V^{{\mu }^{\mathrm{\prime }}}$, since $\mu \mathrm{\ne }{\mu }^{\mathrm{\prime }}$ and the irreducible modules are inequivalent; by step 3.1 the element $e_{\mu }$ therefore separates the two edges.

Separation of edges with the same predecessor. Let $(\mu ,\lambda ),(\mu ,{\lambda }^{\mathrm{\prime }})\in E$ with $\lambda \mathrm{\ne }{\lambda }^{\mathrm{\prime }}$; write $\lambda =\mu +x$, ${\lambda }^{\mathrm{\prime }}=\mu +x^{\mathrm{\prime }}$ with distinct addable nodes $x\mathrm{\ne }{x}^{\mathrm{\prime }}$ of $\mu$. By [F4] their contents differ, $c(x)\mathrm{\ne }c(x^{\mathrm{\prime }})$, so by step 3.1 the element $X_n$ assumes the distinct values $c(x),c(x^{\mathrm{\prime }})$ on the two edges and separates them.

A unital subalgebra of ${\mathbb{C}}^E$ that separates the points of the finite set $E$ is all of ${\mathbb{C}}^E$: indeed, for each $e\in E$ and each $e^{\mathrm{\prime }}\mathrm{\ne }e$ separation gives $a_{e^{\mathrm{\prime }}}\in A$ with $a_{e^{\mathrm{\prime }}}(e)\mathrm{\ne }{a}_{e^{\mathrm{\prime }}}(e^{\mathrm{\prime }})$, and then $b_e:={\prod }_{e^{\mathrm{\prime }}\mathrm{\ne }e}(a_{e^{\mathrm{\prime }}}-a_{e^{\mathrm{\prime }}}(e^{\mathrm{\prime }})\cdot 1)$ lies in $A$, satisfies $b_e(e)\mathrm{\ne }0$ and $b_e(e^{\mathrm{\prime }\mathrm{\prime }})=0$ for $e^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\ne }e$, and is therefore a nonzero multiple of the standard basis idempotent ${\delta }_e$; hence every ${\delta }_e\in A$ and $A={\mathbb{C}}^E$.

By steps 4.1 and 4.2 the subalgebra $A$ of step 3.1 separates every pair of distinct edges, so step 5.1 gives $A={\mathbb{C}}^E$. Since the evaluation map of step 2.1 is an isomorphism, the preimage of ${\mathbb{C}}^E$ is the whole centralizer; hence $⟨Z(\mathbb{C}[S_{n-1}]),X_n⟩=Z(\mathbb{C}[S_n],\mathbb{C}[S_{n-1}])$.

Every central element of $\mathbb{C}[S_n]$ commutes with $\mathbb{C}[S_{n-1}]$, so $Z(\mathbb{C}[S_n])\subseteq Z(\mathbb{C}[S_n],\mathbb{C}[S_{n-1}])=⟨Z(\mathbb{C}[S_{n-1}]),X_n⟩$ by step 6.1.