Background/Motivation
Let $R$ be a commutative ring with unit. If $G$ is a finite (or in general, discrete) group, let $R[G]$ be the group $R$-algebra associated to $G$. The isomorphism problem for group rings asks for condiions on groups $G$ and $H$ such that $R[G]\simeq R[H]$.
Group rings are not a complete invariant, even of finite groups; in a 2001 Annals paper, Hertweck discovered two finite groups $G$ and $H$ with $\mathbb{Z}[G]\simeq \mathbb{Z}[H]$ and $G\not\simeq H$. In general, group algebras over a field are even weaker invariants; for example, if $G$ and $H$ are any two finite abelian groups of order $n$, then $\mathbb{C}[G]\simeq \mathbb{C}[H]\simeq \mathbb{C}^n$, by e.g. the Chinese Remainder Theorem or Artin-Wedderburn.
I am curious about a slight strengthening of the isomorphism problem for group rings. Namely, if $(S, +, \cdot)$ is a (not necessarily commutative) ring with unit, let the opposite ring $S^{op}=(S, +, \times)$ be the ring whose underlying Abelian group under addition is the same as that of $S$, but with the multiplicative structure reversed, i.e. $a\times b=b\cdot a$; the formation of the opposite is clearly functorial. Note that if $R[G]$ is a group ring, it is naturally isomorphic to its opposite through the map $\phi_G: g\mapsto g^{-1}$.
The Problem
Now if $G, H$ are groups and $\psi: R[G]\to R[H]$ is an isomorphism of group rings, we may ask if it is compatible with the formation of the opposite ring---that is, does $\phi_H\circ \psi=\psi^{op}\circ \phi_G$? Say that $G, H$ have strongly isomorphic group rings if such a $\psi$ exists.
What is known about groups with strongly isomorphic group rings over commutative rings $R$? Are there non-isomorphic finite groups $G, H$ with $\mathbb{Z}[G]$ strongly isomorphic to $\mathbb{Z}[H]$, for example? More weakly, when is $\mathbb{C}[G]\simeq \mathbb{C}[H]$?
Strong Isomorphism is Strong
Just to convince you that strong group ring isomorphism is in fact a stronger condition that group ring isomorphism, note that $\mathbb{C}[\mathbb{Z}/4\mathbb{Z}]$ is isomorphic to $\mathbb{C}[\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}]$ but not strongly isomorphic. This is because $\phi_{\mathbb{Z}/4\mathbb{Z}}$ is not the identity, but $\phi_{\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}}$ is the identity on the underlying set of $\mathbb{C}[\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}]$.
