Reversal anti-involution commutes with the Hecke bar

The reversal map descends. On the free associative $A$-algebra, fix every coefficient and generator and reverse each word; this defines an $A$-linear anti-homomorphism. It sends each quadratic relator to itself, each adjacent braid relator to itself because both sides are palindromes, and each distant commutation relator to its negative. Hence it preserves the defining ideal and descends to an $A$-linear anti-homomorphism $\mathrm{♭}$ of $H_v(n)$.

It is an involution with the required formula. Reversing twice fixes every word, so ${\mathrm{♭}}^2=\mathrm{i}\mathrm{d}$ and $\mathrm{♭}$ is an anti-automorphism. If $w=s_{i_1}\cdots s_{i_k}$ is reduced, then $w^{-1}=s_{i_k}\cdots s_{i_1}$ is reduced and $\mathrm{♭}(H_w)=H_{s_{i_k}}\cdots H_{s_{i_1}}=H_{w^{-1}}$. The empty word gives $\mathrm{♭}(H_{\mathrm{i}\mathrm{d}})=H_{\mathrm{i}\mathrm{d}}$.

It commutes with bar. Put $\lambda =v^{-1}-v$. On coefficients, $\mathrm{♭}(\bar{v})=\mathrm{♭}(v^{-1})=v^{-1}=\overline{\mathrm{♭}(v)}$. On a generator, $\mathrm{♭}(\overline{H_{s_i}})=\mathrm{♭}(H_{s_i}-\lambda )=H_{s_i}-\lambda =H_{s_i}^{-1}=\overline{\mathrm{♭}(H_{s_i})}$. Both composites of $\mathrm{♭}$ and bar are semilinear anti-homomorphisms, so agreement on coefficients and generators from the presentation proves that $\mathrm{♭}$ and bar commute on all of $H_v(n)$. No choice principle is used.