In Markl, Schnider, and Stasheff's Operads in algebra, topology, and physics, they give the observation $\mathfrak{s}^{-1} \mathcal{E}nd_V \cong \mathcal{E}nd_{V[-1]}$ (lemma 3.16) as motivation for why the operadic suspension involves the signum representation. I'm having some sign issues when following their calculation for this. The relevant step troubling me is $$ \operatorname{Hom}(((V[-1])^{\otimes n})^j, (V[-1])^{i+j}) \cong \operatorname{Hom}((V^{\otimes n})^{j-n}, V^{i+j-1}) \otimes \operatorname{sgn}_n $$ where $V$ is a differential graded algebra and $\operatorname{sgn}_n$ is the signum representation of $\Sigma_n$ concentrated in degree 0. Here, the symmetric group $\Sigma_n$ acts on $n$-fold tensor products by permuting the tensor factors.
I think this isomorphism ought to follow from $V[-1]^{\otimes n} \cong V^{\otimes n}[-n] \otimes \operatorname{sgn}_n$, and I will be convinced if it works when $n = 2$, but I'm having difficulty showing
$$V[-1] \otimes V[-1] \cong (V \otimes V)[-2] \otimes \operatorname{sgn}_2$$
as $\Sigma_2$-dgas. The isomorphism needs to be $\Sigma_2$-equivariant, hence the need for $\operatorname{sgn}_2$.
Some ideas that I had: I think the issue is that the isomorphism is not the "obvious" one, i.e., not $v \otimes w \mapsto v \otimes w$. In fact, I can identify two places where signs might need to be adjusted. First, we have the Koszul sign rule for interchange. Let $v \in V^{a-1}$ and $w \in V^{b-1}$. We can think of $v \otimes w$ as an element in $V[-1]^a \otimes V[-1]^b$, and we have: $$w \otimes v = (-1)^{ab} v \otimes w.$$
However, the "same" element $v \otimes w \in V^{a-1} \otimes V^{b-1} \subset V \otimes V$ satisfies a different relation: $$w \otimes v = (-1)^{(a-1)(b-1)} v \otimes w.$$
Somehow the isomorphism ought to send the former relation to the latter, and the signum representation may be needed, since we are interchanging the two tensor factors. But I can't quite see how.
A different place where signs appear is the definition of the differential. In $V[-1] \otimes V[-1]$, with $v, w$ as above, we have $$d(v \otimes w) = dv \otimes w + (-1)^a v \otimes dw.$$
But in $V \otimes V$, we have $$d(v \otimes w) = dv \otimes w + (-1)^{a-1} v \otimes dw.$$
Can someone help me out? I think that I'm misunderstanding sign conventions horribly.