V.I. Arnold proved that the image of the hyperelliptic group in the mod 2 symplectic group (ie, the image of $HMod_g\to Mod_g\to Sp_{2g}(\mathbb Z)\to Sp_{2g}(\mathbb F_2)$) is the symmetric group on the branch points: $HMod_g\twoheadrightarrow Mod_{0,2g+2}\twoheadrightarrow S_{2g+2}\hookrightarrow Sp_{2g}(\mathbb F_2)$. For $g=1,2$, this inclusion of finite groups is an isomorphism, but not for $g\geqslant 3$. Moreover, for $g=3$, not every element is conjugate into the symmetric group. In particular, there are elements of order 9 in $Sp_6(\mathbb F_2)$ but not in $S_8$. This is an exercise in linear algebra which I messed up when preparing this answer, so me rely on the symmetry Dan mentioned, which is thus a finite element of the mapping class group, and thus retains its order in the coprime symplectic group. Thus not only does this finite element of the mapping class group not commute with any hyperelliptic element, no element which reduces to it in $Sp_6(\mathbb F_2)$ commutes with any hyperelliptic element.
(Linear algebra: given a Galois extensions $K\subset M$ which is a composite of two Galois extensions $M=L_1L_2$ of degrees $g$ ($L_1/K$) and $2$ ($L_2/K$); put a symplectic form on $M/K$ so that $U_1(M/L_1)\subset U_g(L_2/K)\subset Sp_{2g}(K)$. Ultimately $K=\mathbb F_2$, $L_1=\mathbb F_8$, and $L_2=\mathbb F_4$; so $U_1(F_{64}/F_{8})=\mu_9\subset Sp_6(\mathbb F_2)$.)
