5 deleted 65 characters in body

(I have rewritten parts to respond to potential objections, since someone disliked this answer. I didn't change the thrust of my discussion, but I did alter some places where I might have been slightly too glib. If I missed something, please feel free to correct me in the comments!)

The situation is even better than that! Suppose given an $E_1$-algebra $A$ of a presentable symmetric monoidal ∞-category $\mathcal{C}$.

Call an $E_n$-monoidal structures on the ∞-category $\mathbf{Mod}(A)$ of left $A$-modules allowable if $A$ is the unit and the right action of $\mathcal{C}$ on $\mathbf{Mod}(A)$ is compatible with the $E_n$ monoidal structure, so that $\mathbf{Mod}(A)$ is an $E_n$-$\mathcal{C}$-algebra. Then the space of allowable $E_n$-monoidal structures is equivalent to the space of $E_{n+1}$-algebra structures on $A$ itself, compatible with the extant $E_1$ structure on $A$. (This is even true when $n=0$, if one takes an $E_0$-monoidal category to mean a category with a distinguished object.) The object $A$, regarded as the unit $A$-module, admits an $E_n$-algebra structure that is suitably compatible with the $E_1$ structure an $A$. [Reference: Jacob Lurie, DAG VI, Corollary 2.3.15.]

Let's sketch a proof of this claim in the case Peter mentions. Suppose $A$ is a monoid is a presentable symmetric monoidal category $(\mathbf{C},\otimes)$. Suppose $\mathbf{Mod}(A)$ admits a monoidal structure (not even a priori symmetric!) in which $A$, regarded as a left $A$-module, is the unit. I claim that $A$ is a commutative monoid. Consider the monoid object $\mathrm{End}(A)$ of endomorphisms of $A$ as a left $A$-module; the Eckmann-Hilton argument described below applies to the operations of tensoring and composing to give $\mathrm{End}(A)$ the structure of a commutative monoid object. The multiplication on $A$ yields an isomorphism of monoids $A\simeq\mathrm{End}(A)$.

In the case you mention, the result amounts to the original Eckmann-Hilton, as follows. If $X$ admits magma structures $\circ$ and $\star$ with the same unit (Below, Tom Leinster points out that I only have to assume that each has a unit, and it will follow that the units are the same. He's right, of course.) with the property that

$$(a\circ b)\star(c\circ d)=(a\star c)\circ(b\star d)$$

for any $a,b,c,d\in X$, then (1) the magma structures $\circ$ and $\star$ coincide; (2) the product $\circ$ is associative; and (3) the product $\circ$ is commutative. That is, a unital magma in unital magmas is a commutative monoid.

EDIT (Harry): I just fixed a little problem with the LaTeX.

4 Fixed LaTeX

(I have rewritten parts to respond to potential objections, since someone disliked this answer. I didn't change the thrust of my discussion, but I did alter some places where I might have been slightly too glib. If I missed something, please feel free to correct me in the comments!)

The situation is even better than that! Suppose given an $E_1$-algebra E_1$-algebra$A$of a presentable symmetric monoidal ∞-category $\mathcal{C}$. \mathcal{C}$.

Call an $E_n$-monoidal E_n$-monoidal structures on the ∞-category $\mathbf{Mod}(A)$ of left$A$-modules allowable if $A$ is the unit and the right action of$\mathcal{C}$on$\mathbf{Mod}(A)$is compatible with the$E_n$monoidal structure, so that$\mathbf{Mod}(A)$is an$E_n$-$\mathcal{C}$-algebra. Then the space of allowable$E_n$-monoidal structures is equivalent to the space of$E_{n+1}$-algebra structures on$A$itself, compatible with the extant$E_1$structure on$A$. (This is even true when$n=0$, if one takes an$E_0$-monoidal category to mean a category with a distinguished object.) The object$A$, regarded as the unit$A$-module, admits an$E_n$-algebra structure that is suitably compatible with the$E_1$structure an$A$. [Reference: Jacob Lurie, DAG VI, Corollary 2.3.15.] Let's sketch a proof of this claim in the case Peter mentions. Suppose$A$is a monoid is a presentable symmetric monoidal category$(\mathbf{C},\otimes)$. Suppose$\mathbf{Mod}(A)$admits a monoidal structure (not even a priori symmetric!) in which$A$, regarded as a left$A$-module, is the unit. I claim that$A$is a commutative monoid. Consider the monoid object$\mathrm{End}(A)$of endomorphisms of$A$as a left$A$-module; the Eckmann-Hilton argument described below applies to the operations of tensoring and composing to give$\mathrm{End}(A)$the structure of a commutative monoid object. The multiplication on$A$yields an isomorphism of monoids$A\simeq\mathrm{End}(A)$. In the case you mention, the result amounts to the original Eckmann-Hilton, as follows. If$X$admits magma structures$\circ$and$\star$with the same unit (Below, Tom Leinster points out that I only have to assume that each has a unit, and it will follow that the units are the same. He's right, of course.) with the property that $$(a\circ b)\star(c\circ d)=(a\star c)\circ(b\star d)$$ for any$a,b,c,d\in X$, then (1) the magma structures$\circ$and$\star$coincide; (2) the product$\circ$is associative; and (3) the product$\circ$is commutative. That is, a unital magma in unital magmas is a commutative monoid. EDIT (Harry): I just fixed a little problem with the LaTeX. 3 one last edit (I have rewritten parts to respond to potential objections, since someone disliked this answer. I didn't change the thrust of my discussion, but I did alter some places where I might have been slightly too glib. If I missed something, please feel free to correct me in the comments!) The situation is even better than that! Given Suppose given an$E_1$-algebra$A$of a presentable symmetric monoidal ∞-category, the space of ∞-category$\mathcal{C}$. Call an$E_n$-monoidal structures on the ∞-category$\mathbf{Mod}(A)$of left$A$-modules wherein allowable if$A$is the unit and the right action of$\mathcal{C}$on$\mathbf{Mod}(A)$is compatible with the$E_n$monoidal structure, so that$\mathbf{Mod}(A)$is an$E_n$-$\mathcal{C}$-algebra. Then the space of allowable$E_n$-monoidal structures is equivalent to the space of$E_{n+1}$-algebra structures on$A$itself, compatible with the extant$E_1$structure on$A$. (This is even true when$n=0$, if one takes an$E_0$-monoidal category to mean a category with a distinguished object.) The object$A$, regarded as the unit$A$-module, admits an$E_n$-algebra structure that is suitably compatible with the$E_1$structure an$A$. [Reference: Jacob Lurie, DAG VI, Corollary 2.3.15.] Let's sketch a proof of this claim in the case Peter mentions. Suppose$A$is a monoid is a presentable symmetric monoidal category$(\mathbf{C},\otimes)$. Suppose$\mathbf{Mod}(A)$admits a monoidal structure (not even a priori symmetric!) in which$A$, regarded as a left$A$-module, is the unit. I claim that$A$is a commutative monoid. Consider the monoid object$\mathrm{End}(A)$of endomorphisms of$A$as a left$A$-module; the Eckmann-Hilton argument described below applies to the operations of tensoring and composing to give$\mathrm{End}(A)$the structure of a commutative monoid object. The multiplication on$A$yields an isomorphism of monoids$A\simeq\mathrm{End}(A)$. In the case you mention, the result amounts to the original Eckmann-Hilton, as follows. If$X$admits magma structures$\circ$and$\star$with the same unit (Below, Tom Leinster points out that I only have to assume that each has a unit, and it will follow that the units are the same. He's right, of course.) with the property that $$(a\circ b)\star(c\circ d)=(a\star c)\circ(b\star d)$$ for any$a,b,c,d\in X$, then (1) the magma structures$\circ$and$\star$coincide; (2) the product$\circ$is associative; and (3) the product$\circ\$ is commutative. That is, a unital magma in unital magmas is a commutative monoid.

2 rewritten parts to respond to potential objections
1