The only possibility is that there is exactly one Sylow 13-subgroup, which must be $H$. Since there is only one such subgroup, it must be normal in $G$.
Now, let $h \in H$ and $g \in G$. Since $H$ is normal in $G$, we have $gHg^{-1} = H$. Thus, $ghg^{-1} \in H$
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