= Solution
Choose coordinates $(q_1,\ldots,q_n,p_1,\ldots,p_n)$ with <standard symplectic form> $\omega_0=\sum_jdq_j\wedge dp_j$, and regard the given sphere $S$ as lying in $\{p_n=0\}$. With $\iota_{X_H}\omega_0=-dH$, the linear <Hamiltonian function> $H_\delta=\delta q_n$ has <Hamiltonian vector field> $\delta\partial_{p_n}$ and translates in the missing coordinate.
To make this a permissible compactly supported map, take a smooth cutoff $\chi$ equal to one on an open neighbourhood of the fixed compact swept set
$$
\{z+s e_{p_n}:z\in S,\ |s|\leq1\}.
$$
Choose $\chi$ with compact support and let $R=\sup_{\operatorname{supp}\chi}|q_n|<\infty$. For $0<\delta<1$ put $G_\delta=\delta q_n\chi$. On the whole swept region the cutoff is locally constant, so its derivatives vanish and $X_{G_\delta}=\delta\partial_{p_n}$. Thus the flow carries every $z\in S$ to $z+t\delta e_{p_n}$ for $0\leq t\leq1$.
The time-one image lies in $\{p_n=\delta\}$ and is disjoint from the original sphere. Its <Hofer metric from the spatial supremum norm> obeys
$$
\rho_\infty(\mathrm{id},\phi_{G_\delta}^1)\leq\|G_\delta\|_\infty\leq R\delta.
$$
In the oscillation convention the upper bound is $2R\delta$. Since $\delta$ can approach zero while the cutoff and $R$ stay fixed,
$$
\boxed{e(S)=0}.
$$
This proves the <zero displacement energy for a compact set in a hyperplane>; it does not assert that any one nonidentity displacing map has zero norm.
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