TITLE: A-infinity structure of E-infinity algebras
QUESTION [5 upvotes]: 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 $x^2=0$.

REPLY [2 votes]: as Fernando points out in his comment,
  this is too much to expect in general.
  The A∞-algebra structure on HA is not
  unique

There is a unique $A_\infty$-algebra structure on $H(A)$ such that 
$H(A)$ and $A$ are weakly equivalent (i.e. $A_\infty$-quasi-isomorphic). 
About your question, in characteristic 0 you could replace your $E_\infty$-algebra by a weakly equivalent DG commutative algebra... then on its cohomology you would get a $C_\infty$-structure (which is a strictly commutative $A_\infty$-structure). 
In positive characteristic this is more difficult. But there is still the possibility to deal with divided power algebras (see e.g. http://math.univ-lille1.fr/~fresse/PartitionHomology.pdf on page 18 for a hint). 
I know this is not really an answer to your question. But I hope to can help.