I'm going to write up Mark Wildon's proof as I understand it. As in the standard proof, we start by showing the Key Lemma that the centralizer of $k[S_n]$ is linearly spanned by $GL(V)$. Decompose $V^{\otimes n}$ into $S_n$-irreps, and let $\alpha$ be an endomorphism of $V^{\otimes n}$ commuting with $S_n$. GL(V)$. For each irrep$U$of$S_n$, let$U_1$, ...,$U_a$be the occurrences of$U$in$V^{\otimes n}$. For any$U_i$, consider the endomorphism of$V^{\otimes n}$which acts by$1$on$U_i$and on$0$on all of the other summands of$V^{\otimes n}$. This commutes with$k[S_n]$so, by the Key Lemma it is a linear combination of maps in$GL(V)$. Hence$\alpha$commutes with it, which means that$\alpha$takes$U_i$to$U_i$by some map$\alpha_i$. Consider the endomorphism of$V^{\otimes n}$which takes$U_i$to$U_j$by an$S_n$-equivariant endomorphism and acts by$0$on every other summand of$V^{\otimes n}$. This commutes with$k[S_n]$so, by the Key Lemma it is a linear combination of maps in$GL(V)$. Hence$\alpha$commutes with it, which means that$\alpha_i = \alpha_j$. (Abusing equals to mean "is taken to the other along the isomorphism$U_i \to U_j$, which is unique up to scalar".) Write$\alpha(U)$for the common value of$\alpha_1$,$\alpha_2$, ...,$\alpha_a$. There are now two ways to finish the proof. Standard Argument: By Maschke and Artin-Wedderburn, there is an element in$k[S_n]$which acts on each irrep$U$by$\alpha(U)$. This element of$k[S_n]$induces$\alpha$. Mark Wildon's Argument: Let$V \subset W$. We will show that we can extended$\alpha$to an endomorphism$\beta$of$W^{\otimes n}$which commutes with$GL(W)$. Decompose$W^{\otimes n}$into$S_n$irreps, so that the previous decomposition of$V^{\otimes n}$occurs as a subset of the summands. Let the occurrences of$U$be$U_1 \oplus U_2 \cdots \oplus U_a \oplus \cdots \oplus U_b$. Define a linear map$\beta$\beta:W^{\otimes n}\to W^{\otimes n}$ to acts act on all of the $U_i$ by $\alpha(U)$ or, if $a=0$ so that $\alpha(U)$ is undefined, define $\beta$ to act on the $\alpha_i$ U_i$by$0$. We claim that$\beta$commutes with$GL(W)$. Proof: Any element of$GL(W)$commutes with$k[S_n]$. So (by Schur's lemma), it can only map$U_i$to a linear combination of other$U_j$'s, and the component of$\alpha$mapping$U_i$to$U_j$is a scalar multiple of the standard isomorphism. Clearly,$\beta$commutes with any map of this form. Now, by my argument in the original post, take$\dim W \geq n$to see that$\beta$is induced by an element of$S_n$. Then$\alpha$is also induced by this element of$S_n$. 2 added 2 characters in body I'm going to write up Mark Widon's Wildon's proof as I understand it. As in the standard proof, we start by showing the Key Lemma that the centralizer of$k[S_n]$is linearly spanned by$GL(V)$. Decompose$V^{\otimes n}$into$S_n$-irreps, and let$\alpha$be an endomorphism of$V^{\otimes n}$commuting with$S_n$. For each irrep$U$of$S_n$, let$U_1$, ...,$U_a$be the occurrences of$U$in$V^{\otimes n}$. For any$U_i$, consider the endomorphism of$V^{\otimes n}$which acts by$1$on$U_i$and on$0$on all of the other summands of$V^{\otimes n}$. This commutes with$k[S_n]$so, by the Key Lemma it is a linear combination of maps in$GL(V)$. Hence$\alpha$commutes with it, which means that$\alpha$takes$U_i$to$U_i$by some map$\alpha_i$. Consider the endomorphism of$V^{\otimes n}$which takes$U_i$to$U_j$by an$S_n$-equivariant endomorphism and acts by$0$on every other summand of$V^{\otimes n}$. This commutes with$k[S_n]$so, by the Key Lemma it is a linear combination of maps in$GL(V)$. Hence$\alpha$commutes with it, which means that$\alpha_i = \alpha_j$. (Abusing equals to mean "is taken to the other along the isomorphism$U_i \to U_j$, which is unique up to scalar".) Write$\alpha(U)$for the common value of$\alpha_1$,$\alpha_2$, ...,$\alpha_a$. There are now two ways to finish the proof. Standard Argument: By Maschke and Artin-Wedderburn, there is an element in$k[S_n]$which acts on each irrep$U$by$\alpha(U)$. This element of$k[S_n]$induces$\alpha$. Mark Widon's Wildon's Argument: Let$V \subset W$. We will show that we can extended$\alpha$to an endomorphism$\beta$of$W^{\otimes n}$which commutes with$GL(W)$. Decompose$W^{\otimes n}$into$S_n$irreps, so that the previous decomposition of$V^{\otimes n}$occurs as a subset of the summands. Let the occurrences of$U$be$U_1 \oplus U_2 \cdots \oplus U_a \oplus \cdots \oplus U_b$. Define$\beta$to acts on all of the$U_i$by$\alpha(U)$or, if$a=0$so that$\alpha(U)$is undefined, define$\beta$to act on the$\alpha_i$by$0$. We claim that$\beta$commutes with$GL(W)$. Proof: Any element of$GL(W)$commutes with$k[S_n]$. So (by Schur's lemma), it can only map$U_i$to a linear combination of other$U_j$'s, and the component of$\alpha$mapping$U_i$to$U_j$is a scalar multiple of the standard isomorphism. Clearly,$\beta$commutes with any map of this form. Now, by my argument in the original post, take$\dim W \geq n$to see that$\beta$is induced by an element of$S_n$. Then$\alpha$is also induced by this element of$S_n$. 1 [made Community Wiki] I'm going to write up Mark Widon's proof as I understand it. As in the standard proof, we start by showing the Key Lemma that the centralizer of$k[S_n]$is linearly spanned by$GL(V)$. Decompose$V^{\otimes n}$into$S_n$-irreps, and let$\alpha$be an endomorphism of$V^{\otimes n}$commuting with$S_n$. For each irrep$U$of$S_n$, let$U_1$, ...,$U_a$be the occurrences of$U$in$V^{\otimes n}$. For any$U_i$, consider the endomorphism of$V^{\otimes n}$which acts by$1$on$U_i$and on$0$on all of the other summands of$V^{\otimes n}$. This commutes with$k[S_n]$so, by the Key Lemma it is a linear combination of maps in$GL(V)$. Hence$\alpha$commutes with it, which means that$\alpha$takes$U_i$to$U_i$by some map$\alpha_i$. Consider the endomorphism of$V^{\otimes n}$which takes$U_i$to$U_j$by an$S_n$-equivariant endomorphism and acts by$0$on every other summand of$V^{\otimes n}$. This commutes with$k[S_n]$so, by the Key Lemma it is a linear combination of maps in$GL(V)$. Hence$\alpha$commutes with it, which means that$\alpha_i = \alpha_j$. (Abusing equals to mean "is taken to the other along the isomorphism$U_i \to U_j$, which is unique up to scalar".) Write$\alpha(U)$for the common value of$\alpha_1$,$\alpha_2$, ...,$\alpha_a$. There are now two ways to finish the proof. Standard Argument: By Maschke and Artin-Wedderburn, there is an element in$k[S_n]$which acts on each irrep$U$by$\alpha(U)$. This element of$k[S_n]$induces$\alpha$. Mark Widon's Argument: Let$V \subset W$. We will show that we can extended$\alpha$to an endomorphism$\beta$of$W^{\otimes n}$which commutes with$GL(W)$. Decompose$W^{\otimes n}$into$S_n$irreps, so that the previous decomposition of$V^{\otimes n}$occurs as a subset of the summands. Let the occurrences of$U$be$U_1 \oplus U_2 \cdots \oplus U_a \oplus \cdots \oplus U_b$. Define$\beta$to acts on all of the$U_i$by$\alpha(U)$or, if$a=0$so that$\alpha(U)$is undefined, define$\beta$to act on the$\alpha_i$by$0$. We claim that$\beta$commutes with$GL(W)$. Proof: Any element of$GL(W)$commutes with$k[S_n]$. So (by Schur's lemma), it can only map$U_i$to a linear combination of other$U_j$'s, and the component of$\alpha$mapping$U_i$to$U_j$is a scalar multiple of the standard isomorphism. Clearly,$\beta$commutes with any map of this form. Now, by my argument in the original post, take$\dim W \geq n$to see that$\beta$is induced by an element of$S_n$. Then$\alpha$is also induced by this element of$S_n\$.