Assume $v:H^{\otimes m}\to H^{\otimes m}$ is a linear operator on the $n^{\text{th}}$ tensor power of a vector space $H$. For each permutation $p$ on the $m$-element set define the linear operator $p:H^{\otimes m}\to H^{\otimes m}$ by $p(x_1\otimes \dots\otimes x_n)=(x_{p(1)}\otimes \dots\otimes x_{p(m)})$.
Now fix $n > m$ and let $1\leq j_1,\dots,j_m\leq n$ be two-by-two different and define the linear operator $V_{j_1,\dots,j_m}:H^{\otimes n}\to H^{\otimes n}$ by $V_{j_1,\dots,j_m}(x_1\otimes\dots\otimes x_n) = k_{j_1,\dots,j_m}^{-1}\circ (v\otimes Id \otimes \dots \otimes Id)\circ k_{j_1,\dots,j_m}$, where $k_{j_1,\dots,j_m}$ is the permutation operator that takes the elements in positions $j_1,\dots,j_m$ and puts them in the first $m$ positions in the same order.
Proposition: If $v$ commutes with each permutation operator then so does $V_{j_1,\dots,j_m}$.
Thought: Any permutation disjoint with $k_{j_1,\dots,j_m}$ commutes with $V$ because disjoint permutations commute. How can I get the general permutation to commute through?
EDIT: You may additionally assume that $j_i>j_k$ whenever $i>k$.