No.
If a group $G$ of rigid motions operates transitively on $\mathbb R^n$, then also the orbit of the origin is $\mathbb R^n$. But the elements of $O(n)$ all fix $0$, so $G$ must contain all translations.
No.
If a group $G$ of rigid motions operates transitively on $\mathbb R^n$, then also the orbit of the origin is $\mathbb R^n$. But the elements of $O(n)$ all fix $0$, so $G$ must contain all translations.