This is perhaps somewhat related to this question. Fix a field $k$ of characteristic $p>0$. Suppose that $A$ is an $E_\infty$-algebra over $k$. Then $A$ also has an $A_\infty$-algebra structure, and therefore so does its homology $HA$. Its homology is also graded commutative.

I'm looking for an extended version of graded commutativity, a result like this: if $m_n$ is any of the higher multiplications in the $A_\infty$ structure on $HA$, then for any $1 \leq j \leq n$, $$ x m_n(a_1 \otimes \cdots \otimes a_n) = \pm m_n(a_1 \otimes \cdots \otimes x a_j \otimes \cdots \otimes a_n). $$ This is certainly true if $n=2$ by graded commutativity. What about for larger values?

(I'm tempted to tag any question about $E_\infty$-algebras as "commutative algebra", but I suppose that would be misleading...)

Edit: as Fernando points out in his comment, this is too much to expect in general. The $A_\infty$-algebra structure on $HA$ is not unique, so perhaps the right question is, are there conditions on $x$ and the $a_i$ so that, for some choice of $m_n$, $x m_n(\dots) = \dots$?

Along with Fernando's example, another one to consider is the mod $p$ cohomology of a cyclic group of order $p$, with $p$ odd. If $x$ is the generator of $H^1$ and $y$ is the generator of $H^2$, then $m_p(x^{\otimes p}) = \pm y$, so $x m_p(x^{\otimes p}) = \pm xy \neq 0$ while I think $m_p(x^2 \otimes x^{\otimes p-1}) = 0$, since $x^2=0$.