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 $A$?



Opposite of an E2-algebra
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.


