Opposite of an E2-algebra Suppose $C$ is the monoidal $\infty$-category of modules over an $\mathcal{E}_2$-ring spectrum $A$.  Let $C' = C$ as a category, but with opposite monoidal structure to $C$.  Is $C'$ the category of modules over another $\mathcal{E}_2$-ring spectrum $A'$?  Is there standard notation for $A'$ in terms of $A$?
 A: You should call $A'$ the "opposite" of $A$.  The $E_2$-operad has an automorphism given by reflecting in the plane in which the operad is defined (in terms of disks or squares or whatever).  Your algebra $A'$ is the pullback of $A$ along this automorphism.
Does the choice of reflection matter?  Sort of.  The ratio of two reflections is a rotation.  The rotations of the $E_2$ operad are all homotopy-equivalent automorphisms, but there is $\pi_1$ in the automorphism group.  It acts on an $E_2$-algebra $A$ by the algebra automorphism $A \to A$ which is the identity on underlying spaces (or spectra, or whatever) but which is made into an $E_2$-homomorphism by a non-identity homotopy $xy \overset\sim\to xy$.  So which reflection should you pick?  You should pick the one in the direction that corresponds to how you make $\mathrm{Mod}_A$ into a monoidal category.  Said another way, to say the words "$\mathrm{Mod}_A$" at all you had to choose a way to think of $A$ as an $E_1$-algebra.  All such choices are equivalent, of course, but there's homotopy in the space of choices, and making a choice is the same as picking out a line in the plane, which picks out an operad map $E_1 \to E_2$ along which you can pull back $A$.  The reflection you should use is the one that fixes this choice of line.
As an example of all this, replace the word "spectrum" by "strict category".  Then an $E_2$-ring spectrum becomes a braided monoidal category, and the "opposite" is the one where the braiding is replaced by its inverse.
