The diagonal tensor action of $\mathbb{C}[C_2]$ on trivial and sign modules

Define ${\rho }_V(1)={\rho }_W(1)=1$, ${\rho }_V(s)=1$, and ${\rho }_W(s)=-1$, and extend linearly from the basis $\{1,s\}$. These maps preserve products because the only nontrivial basis relation is $s^2=1$ and both $1^2=(-1{)}^2=1$. Thus they are unital representations by [F3], hence left $H$-modules by [F4].

By [F1]–[F2], $H$ is a bialgebra with $\mathrm{\Delta }(1)=1\otimes 1$ and $\mathrm{\Delta }(s)=s\otimes s$. The action in [F5] sends $1$ to the identity and sends $s$ to $-{\mathrm{id}}_{V\otimes W}$, since $s\cdot (v\otimes w)=(s\cdot v)\otimes (s\cdot w)=v\otimes (-w)$. This operator squares to the identity, so the algebra homomorphism $H\to {\mathrm{End}}_{\mathbb{C}}(V\otimes W)$ determined on the basis $1,s$ gives a unital module action and agrees with the coproduct formula on that basis, hence on all of $H$.

The vector-space flip $\tau (v\otimes w)=w\otimes v$ is an isomorphism by [F6]. It is $H$-linear: for $s$, both $\tau (s\cdot (v\otimes w))$ and $s\cdot \tau (v\otimes w)$ equal $-w\otimes v$, while $1$ acts identically. Linearity and the basis $\{1,s\}$ give the intertwining identity for every $h\in H$. This is a symmetric swap for this example, not a claim that every bialgebra-module category is symmetric.