Let $A$ be a finite dimensional semisimple quasitriangular Hopf algebra over $\mathbb{C}$ with universal $R$-matrix denoted by $\mathcal{R}\in A\otimes A$. The Drinfeld element of $A$ is defined as$$u=m(S\otimes\mathrm{id})\mathcal{R}_{21}.$$For finite dimensional semisimple Hopf algebras, $u$ is a central element, i.e., $ux=xu$ for $x\in A$.
A Drinfeld twist for $A$ is an invertible element $J=J^{(1)}\otimes J^{(2)}\in A\otimes A$, with inverse $\bar{J}=\bar{J}^{(1)}\otimes \bar{J}^{(2)}\in A\otimes A$ satisfying\begin{equation} (\Delta\otimes \mathrm{id})(J)J_{12}=(\mathrm{id}\otimes \Delta)(J)J_{23}, \;\; (\varepsilon\otimes \mathrm{id})(J)=(\mathrm{id}\otimes \varepsilon)(J)=1,\;\; J\bar{J}=1\otimes 1,\end{equation}where $\mathrm{id}$ is the identity map of $A$. Given a twist $J$ for $A$, we can define the twisted quasitriangular Hopf algebra$(A^J,\Delta^J,\varepsilon^J,S^J, \mathcal{R}^J)$ as follows: $A^J=A$ as analgebra, the coproduct, counit, antipode, and universal $R$-matrix are determined by\begin{eqnarray}\Delta^J(x)=J^{-1}\Delta(x)J,\quad \varepsilon^J=\varepsilon,\quad S^J(x)=w^{-1}S(x)w,\quad \mathcal{R}^J=J_{21}^{-1}\mathcal{R}J\end{eqnarray}for all $x\in A$, where $w=m(S\otimes \mathrm{id})J=SJ^{(1)}J^{(2)}$ is an invertible element of $A$ with $w^{-1}=\bar{J}^{(2)}S\bar{J}^{(1)}$.
Question: is the Drinfeld element of $A^J$ the same as the Drinfeld element of $A$?
