I asked [this question](https://math.stackexchange.com/questions/3766785/show-that-a-linear-map-commutes-with-a-symmetric-group-action) on MSE but I want to ask it again here with some more context sine it received no answers. In Chapter 3 (Algebra) of the book *Operads in Algebra, Topology and Physics* by Markl, Shnider and Staffesh there is the Lemma 3.16 where the authors state the isomorphism of operads $\mathfrak{s}^{-1}\mathrm{End}\_V\cong \mathrm{End}\_{\Sigma V}$. For the definition of the operadic suspension $\mathfrak{s}$ and other notation please see my [previous question](https://mathoverflow.net/questions/365140/is-operadic-desuspension-inverse-to-operadic-suspension).


But the proof they give is not very explicit, since they give (non-explicit) isomorphisms of graded modules and don't really justify that they are morphisms of operads. I tried to show that and I came across a problem trying to show that the isomorphism commutes with the action of the symmetric group. This is where my MSE questions begins, so I am going to copy it here.


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 $f\mapsto \Sigma^{-1}\circ f\circ\Sigma^{\otimes n}$ which is indeed an isomorphism of graded modules.



> 
> 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.



Detailed proof of $\mathfrak{s}^{-1}\mathrm{End}_V\cong \mathrm{End}_{\Sigma V}$
Your actual Question has nothing to do with operads. Perhaps it is clarifying to consider the following more general setting: let $G$ be a group, $X$ and $Y$ be right $G$-sets, and $f : X \to Y$ be a function. If $g, h \in G$ and $f$ commutes with the actions of $g$ and of $h$ then it commutes with the action of $gh$:
$$f(x) \cdot (gh) = (f(x) \cdot g) \cdot h = f(x \cdot g) \cdot h = f((x \cdot g) \cdot h) = f(x \cdot (gh)).$$
So if $f$ commutes with a set of elements which generate $G$ then it commutes with all elements of $G$.


