= Solution
For the conjugation action, fixed elements have coefficients constant on $G$-conjugacy classes. Therefore \b[$(\mathcal OG)^G=Z(\mathcal OG)$]. More generally an $\mathcal O$-basis of $(\mathcal OG)^H$ consists of sums of the $H$-conjugation orbits on $G$.
For $a\in(\mathcal OG)^H$ define the <relative trace>, or transfer, by
$$
\operatorname{Tr}_H^G(a)=\sum_{g\in G/H}gag^{-1}.
$$
The expression is independent of the coset representatives and is $G$-fixed. If $z\in Z(\mathcal OG)$, then $z\operatorname{Tr}_H^G(a)=\operatorname{Tr}_H^G(za)$; hence its image is a <transfer ideal of conjugation-fixed elements> in the <center of an associative algebra>.
Let $P$ be a Sylow $p$-subgroup of $H$. Trace transitivity gives $\operatorname{Tr}_P^G=\operatorname{Tr}_H^G\operatorname{Tr}_P^H$, proving inclusion of the $P$-image in the $H$-image. Conversely $[H:P]$ is a unit of $\mathcal O$, and for an $H$-fixed element $a$,
$$
\operatorname{Tr}_P^H(a)=[H:P]a.
$$
Thus $\operatorname{Tr}_H^G(a)=\operatorname{Tr}_P^G(a/[H:P])$, proving the reverse inclusion and
$$
\boxed{(\mathcal OG)_H^G=(\mathcal OG)_P^G.}
$$
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