Let $V$ be a graded vector space and $\text{End}_V(n)=\hom(V^{\otimes n},V)$. There is a natural action of the symmetric group $S_n$ on $\text{End}_V(n)$ by permuting the arguments, i.e. if $f\in \text{End}_V(n)$ and $\sigma\in S_n$, $(f\sigma)(v_1\otimes\cdots\otimes v_n)=\varepsilon(\sigma)f(v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)})$, where $\varepsilon(\sigma)$ is the Koszul sign produced by permuting $v_1,\dots,v_n$ via $\sigma$. We can twist this action by the sign of $\sigma$, i.e. we consider the action
$$(f\sigma)(v_1\otimes\cdots\otimes v_n)=(-1)^{\sigma}\varepsilon(\sigma)f(v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)})$$
If $\Sigma V$ is the suspension of $V$, we consider the natural action of the symmetric group on $\text{End}_{\Sigma V}(n)$ (the first I defined, without the twist).
There is a map $\phi:\text{End}_{\Sigma V}(n)\to \text{End}_V(n)$ given by $f\mapsto \Sigma^{-1}\circ f\circ\Sigma^{\otimes n}$.
I need to show that $\phi$ commutes wit the action of the symmetric group, where we have the natural action on the domain and the twisted action on the codomain.
I can show this for transpositions of the form $\sigma=(i\ i+1)$. On the one hand,
$$\phi(f\sigma)(v_1\otimes\cdots\otimes v_n)=(-1)^{\sum_{j=1}^n (n-j)v_j}\Sigma^{-1}\circ (f\sigma)(\Sigma v_1\otimes\cdots\otimes \Sigma v_n)=$$
$$(-1)^{\sum_{j=1}^n (n-j)v_j+(v_i-1)(v_{i+1}-1)}\Sigma^{-1}\circ f(\Sigma v_1\otimes\cdots\otimes\Sigma v_{i+1}\otimes\Sigma v_i\otimes\cdots\otimes \Sigma v_n).$$
On the other hand
$$(\phi(f)\sigma) (v_1\otimes\cdots\otimes v_n)=(-1)^{v_iv_{i+1}-1}\Sigma^{-1}\circ f\circ \Sigma^{\otimes n}(v_1\otimes\cdots\otimes v_{i+1}\otimes v_i\otimes\cdots\otimes v_n)=$$
$$(-1)^{v_iv_{i+1}-1+\sum_{j\neq i,i+1}(n-j)v_j +(n-i-1)v_i+(n-i)v_{i+1}}\Sigma^{-1}\circ f(\Sigma v_1\otimes\cdots\otimes \Sigma v_{i+1}\otimes \Sigma v_i\otimes\cdots\otimes \Sigma v_n).$$
Now I just have to check that the signs are the same. Modulo $2$, the sign of the first map is
$$v_iv_{i+1}+v_i+v_{i+1}-1+\sum_{j=1}^n(n-j)v_j=$$ $$v_iv_{i+1}-1+\sum_{j\neq i,i+1}^n(n-j)v_j+(n-i-1)v_i+(n-i)v_{i+1},$$
which indeed coincide with the sign on the second map.
Question: Since these transpositions generate the symmetric group I feel that I should be able to conclude that the action commutes with $\phi$, but I don't know how to do it.