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{eqnarray} (\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{eqnarray} 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 an algebra, 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 $w^{-1}=\bar{J}^{(1)}S\bar{J}^{(2)}$.
Question: is the Drinfeld element of $A^J$ the same as the Drinfeld element of $A$?
