Suppose $\Sigma_p$ is the symmetric group of order $p!$. Let $V$ be a graded $\mathbb{F}_p$-vector space such that $V_i$ is non-trivial for $i$ arbitrary large at even and order degree.
Claim: $H^j(\Sigma_p, V^{\otimes p}) \neq 0$ for $j$ large, where the action of symmetric group on $V^{\otimes p}$ is by permuting the tensor product. Note that this cohomology is over finite field $\mathbb{F}_p$.
Proof: Since $V$ contains $\mathbb{F}_p$-summand $\mathbb{F}_p[i]$ for $i$ arbitrary large even and odd, $V^{\otimes p}$ contains $\Sigma_p$-summand $\mathbb{F}_p[i]^{\otimes p}$, where this is $\mathbb{F}_p[pi]$ for $i$ even and $sgn[pi]$ for $i$ odd. Because we know by J.P May that
$$H^*(\Sigma_p, \mathbb{F}_p) = \wedge_{\mathbb{F}_p}[v] \otimes \mathbb{F}_p[\beta v],$$ where $v$ is a class of degree $2(p-1)-1$.
This shows that $H^*(\Sigma_p, V^{\otimes p}) \neq 0$ since it contains summand $H^*(\Sigma_p, \mathbb{F}_p[i]^{\otimes p})$.
Something is pretty fishy for my proof here. Does that mean cohomology of symmetric group of order $p!$ over any coefficient module tensor $p$ times is non-trivial?
Some help could be great!