Let $G$ be a (discrete, say) group and $\mathbb K$ a field. The *regular representation* $G^{\mathbb K}$ is the vector space of all functions $G \to \mathbb K$. It is a (left, say) $G$-module: given $g\in G$ and $f: G \to \mathbb K$, the action is $g\cdot f: x \mapsto f(xg)$. Then $G^{\mathbb K}$ is a commutative algebra object in $G\text{-rep}_{\mathbb K}$, the symmetric monoidal category of $\mathbb K$-valued $G$-representations, under pointwise multiplication $f_1f_2: x \to f_1(x)f_2(x)$.

But the pointwise product is not necessarily the only commutative algebra (in $G\text{-rep}$) structure that can be put on $G^{\mathbb K}$. For example, when $\mathbb K = \mathbb R$ and $G = \mathbb Z/2$, as an algebra $G^{\mathbb K} \cong \mathbb R[\epsilon]/(\epsilon^2 = 1)$, with the $G$-action corresponding to conjugation $\epsilon \mapsto -\epsilon$. The same $G$-module supports the algebra structure $\mathbb R[\epsilon]/(\epsilon^2 = -1) = \mathbb C$, which is patently a different algebra.

My question is whether there are any examples with $\mathbb K = \mathbb C$? I.e.:

Does there exist a group $G$ so that there is a commutative algebra object in $G\text{-rep}_{\mathbb C}$ that is isomorphic to $G^{\mathbb C}$ as a representation but not as an algebra?

I believe that any such group must be rather large: in particular, I'm sure that it cannot be finite.