The endomorphism monoid of the unit of a monoidal category is commutative. In particular, a monoid has a monoidal structure as one-object category only if it is commutative. Moreover, in that case monoidal structures are given by $f\otimes g=f\circ g$, a strict associator, and an invertible element, considered as an automorphism $m\colon I\otimes I=I\to I$, corresponding to both unitors $\lambda_I$ and $\rho_I$ (the monoidal axioms follow immediately).

For the proof, coherence requires that $\lambda_I=\rho_I\colon I\otimes I\to I$ for the unit $I$ of a monoidal category (this used to be one of the original coherence axioms until shown redundant).

Combined with the fact that $\lambda_X\colon I\otimes X\to X$ and $\rho_X\colon X\otimes I\to X$ are natural, we have an isomorphism $\lambda_I=\rho_I=m\colon I\otimes I\to I$ such that $m\circ I\otimes f=f\circ m=m\circ f\otimes I$ for each endomorphism $f\colon I\to I$. Since $f=(f\circ m)\circ m^{-1}$, we get that $f*g=m\circ f\otimes g\circ m^{-1}$ is a binary operation on endomorphisms of $I$ with unit $\mathrm{id}_I$ that is homomorphic with respect to composition. Eckmann--Hilton therefore applies to conclude $*$ and $\circ$ coincide and are commutative. In particular, if $I\otimes I=I$, then $f\otimes g=m^{-1}\circ(f\circ g)\circ m=f\circ g$ because $m$ commutes with $(f\circ g)$.

Finally, coherence also requires that $a_{I,I,I}\colon I\otimes I\otimes I\to I\otimes I\otimes I$ be the identity since the latter factors as $m^{-1}\otimes I\circ m^{-1}\circ m\circ m\otimes I\colon I\otimes I\otimes I\to I\otimes I\to I\to I\otimes I\to I\otimes I\otimes I$.

$\newcommand{id}{\operatorname{id}}$

I believe I've resolved my problem.

The 3-cocycle condition says that
$$\omega(g_1g_2,g_3,g_4)\omega(g_1,g_2,g_3g_4)=\omega(g_1,g_2,g_3)\omega(g_1,g_2g_3,g_4)\omega(g_2,g_3,g_4)$$
for all $g_i\in G$.
Set $g_1=g^{-1},g_2=g, g_3=g^{-1},g_4=g$.
The cocycle condition says that
$$\omega(1,g^{-1},g)\omega(g^{-1},g,1)=\omega(g^{-1},g,g^{-1})\omega(g^{-1},1,g)\omega(g,g^{-1},g).$$
Since $\omega$ is normalized, $\omega(g^{-1},1,g)=1$, so the right hand side is what we want to compute.
If we show the left hand side is $1$, we will be done.

The first commutative diagram of EGNO Proposition 2.2.4 says $l_X \otimes \id_Y = l_{X\otimes Y} \circ a_{\mathbf{1},X,Y}$ for all objects $X,Y$, and the second commutative diagram says $r_{X\otimes Y} = (\id_X\otimes r_Y)\circ a_{X,Y,\mathbf{1}}$ for all objects $X,Y$.
Replacing $Y$ with $X$ and $X$ with $X^*$ in the proposition, these equalities give relations
\begin{align}
r_{X^*\otimes X} &=\omega(g^{-1},g,1) \id_{X^*}\otimes r_{X} \\
l_{X^*}\otimes\id_{X} &=\omega(1,g^{-1},g)l_{X^*\otimes X},
\end{align}
but EGNO Exercise 2.3.9 says that $l_Z=r_Z=\id_Z$ for all objects $Z$ if and only if $\omega$ is normalized.
So all morphisms in the above relations pick up factors of $1$.
Hence $\omega(g^{-1},g,1)\omega(1,g^{-1},g)=1$.

The same argument goes through by symmetry by swapping roles $g^{-1}\leftrightarrow g$ and $X^* \leftrightarrow X$.

## Best Answer

It is not true that $\delta_{m'}$ is a left dual of $\delta_m$ unless $m'$ is inverse to $m$ on

bothsides. Indeed, for $\delta_{m'}$ to be left dual to $\delta_m$, you need a map $\delta_{m'}\otimes \delta_m\to 1$ and a map $1\to\delta_m\otimes\delta_{m'}$ that make certain diagrams commute. If $mm'$ and $m'm$ are not both equal to the identity element, then one of these maps will be forced to be $0$, and so the required diagrams will not be able to commute (they will end up saying the identity maps on $\delta_m$ and $\delta_{m'}$ have to be equal to $0$).