In the simply-transitive case, there is the following result.

Theorem. Let $G$ be a simply-transitive group of affine rigid motions acting on a finite dimensional vector space $V$. Then $G$ is a connected solvable group, and there exists an orthogonal decomposition $V=U\oplus W$, where $U\neq0$, and a homomorphism $\varphi:V\to \mathfrak{so}(V)$ such that $\ker\varphi=U$, $\mathrm{im}\,\varphi$ is an Abelian subalgebra of $\mathfrak{so}(V)$ annihilating $W$ and preserving $U$, and \begin{equation} \mathfrak g=\{\tau_v+\varphi(v):v\in V\}, \end{equation} where $\tau_v:V\to V$ denotes translation by $v$.

Proof. $G$ is diffeomorphic to $V$, thus connected and contractible. By The Levi-Malcev theorem, it is solvable.

Consider the homomorphism $\pi:G\to SO(V)$ given by $\pi(g)=g_*$. Then $\mathfrak n=\ker d\pi$ is an ideal of $\mathfrak g$ and $\pi(G)$ is a solvable subgroup of $SO(V)$. The closure $\overline{\pi(G)}$ in $SO(V)$ is a compact solvable group, hence, Abelian. It follows that $\pi(G)$ is Abelian.

The map $\theta:\mathfrak g\to V$, $\theta(X)=X\cdot 0$, is an isomorphism of vector spaces. Let $U=\theta(\mathfrak n)$ and $W=U^\perp$. If $U=0$, then $\mathfrak n=0$ and $d\pi[\mathfrak g]$ is an Abelian subalgebra of $\mathfrak{so}(V)$ of dimension $\dim\mathfrak g=\dim V$. However, the rank of $\mathfrak{so}(V)$ is at most $\frac12\dim V$, so this cannot be. It follows that $U\neq0$.

Set $\varphi=d\pi\circ\theta^{-1}$. Note that $\varphi$ is a homomorphism between the Abelian Lie algebras $V$ and $d\pi[\mathfrak g]$. It is also clear that $\mathfrak g$ has the required form. Of course, $\ker\varphi=\theta(\ker d\pi)=\theta(\mathfrak n)=U$.

Let $u\in U$, $w\in W$. Then $$[\tau_w+\varphi(w),\tau_u]=\tau_{\varphi(w)u} \in\mathfrak g, $$ so $\varphi(w)u\in \ker\varphi=U$. This shows that $\mathrm{im}\,\varphi$ preserves $U$ and, hence, the orthogonal decomposition $ V=U\oplus W$. Finally, if $w$, $w'\in W$, then $[\varphi(w),\varphi(w')]=0$, and so $$ [\tau_w+\varphi(w),\tau_{w'}+\varphi(w')]=\tau_{\varphi(w)w'} -\tau_{\varphi(w')w}=\tau_{\varphi(w)w'-\varphi(w')w} \in\mathfrak g. $$ This shows that $\varphi(w)w'-\varphi(w')w\in W\cap U=0$. Now the trilinear form on $W$ $$ (w,w',w'')\mapsto \langle\varphi(w)w',w''\rangle $$ is symmetric in the first two variables and skew-symmetric in the last two variables, hence, identically zero. This proves that $\varphi(W)W=0$ and completes the proof of the theorem. q.e.d.

In the simply-transitive case, there is the following result.

Theorem. Let $G$ be a simply-transitive group of affine rigid motions acting on a finite dimensional Euclidean space $V$. Then $G$ is a connected solvable group, and there exists an orthogonal decomposition $V=U\oplus W$, where $U\neq0$, and a homomorphism $\varphi:V\to \mathfrak{so}(V)$ such that $\ker\varphi=U$, $\mathrm{im}\,\varphi$ is an Abelian subalgebra of $\mathfrak{so}(V)$ annihilating $W$ and preserving $U$, and \begin{equation} \mathfrak g=\{\tau_v+\varphi(v):v\in V\}, \end{equation} where $\tau_v:V\to V$ denotes translation by $v$.

Proof. $G$ is diffeomorphic to $V$, thus connected and contractible. By The Levi-Malcev theorem, it is solvable.

Consider the homomorphism $\pi:G\to SO(V)$ given by $\pi(g)=g_*$. Then $\mathfrak n=\ker d\pi$ is an ideal of $\mathfrak g$ and $\pi(G)$ is a solvable subgroup of $SO(V)$. The closure $\overline{\pi(G)}$ in $SO(V)$ is a compact solvable group, hence, Abelian. It follows that $\pi(G)$ is Abelian.

The map $\theta:\mathfrak g\to V$, $\theta(X)=X\cdot 0$, is an isomorphism of vector spaces. Let $U=\theta(\mathfrak n)$ and $W=U^\perp$. If $U=0$, then $\mathfrak n=0$ and $d\pi[\mathfrak g]$ is an Abelian subalgebra of $\mathfrak{so}(V)$ of dimension $\dim\mathfrak g=\dim V$. However, the rank of $\mathfrak{so}(V)$ is at most $\frac12\dim V$, so this cannot be. It follows that $U\neq0$.

Set $\varphi=d\pi\circ\theta^{-1}$. Note that $\varphi$ is a homomorphism between the Abelian Lie algebras $V$ and $d\pi[\mathfrak g]$. It is also clear that $\mathfrak g$ has the required form. Of course, $\ker\varphi=\theta(\ker d\pi)=\theta(\mathfrak n)=U$.

Let $u\in U$, $w\in W$. Then $$[\tau_w+\varphi(w),\tau_u]=\tau_{\varphi(w)u} \in\mathfrak g, $$ so $\varphi(w)u\in \ker\varphi=U$. This shows that $\mathrm{im}\,\varphi$ preserves $U$ and, hence, the orthogonal decomposition $ V=U\oplus W$. Finally, if $w$, $w'\in W$, then $[\varphi(w),\varphi(w')]=0$, and so $$ [\tau_w+\varphi(w),\tau_{w'}+\varphi(w')]=\tau_{\varphi(w)w'} -\tau_{\varphi(w')w}=\tau_{\varphi(w)w'-\varphi(w')w} \in\mathfrak g. $$ This shows that $\varphi(w)w'-\varphi(w')w\in W\cap U=0$. Now the trilinear form on $W$ $$ (w,w',w'')\mapsto \langle\varphi(w)w',w''\rangle $$ is symmetric in the first two variables and skew-symmetric in the last two variables, hence, identically zero. This proves that $\varphi(W)W=0$ and completes the proof of the theorem. q.e.d.

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.

In the simply-transitive case, there is the following result.

Theorem. Let $G$ be a simply-transitive group of affine rigid motions acting on a finite dimensional vector space $V$. Then $G$ is a connected solvable group, and there exists an orthogonal decomposition $V=U\oplus W$, where $U\neq0$, and a homomorphism $\varphi:V\to \mathfrak{so}(V)$ such that $\ker\varphi=U$, $\mathrm{im}\,\varphi$ is an Abelian subalgebra of $\mathfrak{so}(V)$ annihilating $W$ and preserving $U$, and \begin{equation} \mathfrak g=\{\tau_v+\varphi(v):v\in V\}, \end{equation} where $\tau_v:V\to V$ denotes translation by $v$.

Proof. $G$ is diffeomorphic to $V$, thus connected and contractible. By The Levi-Malcev theorem, it is solvable.

Consider the homomorphism $\pi:G\to SO(V)$ given by $\pi(g)=g_*$. Then $\mathfrak n=\ker d\pi$ is an ideal of $\mathfrak g$ and $\pi(G)$ is a solvable subgroup of $SO(V)$. The closure $\overline{\pi(G)}$ in $SO(V)$ is a compact solvable group, hence, Abelian. It follows that $\pi(G)$ is Abelian.

The map $\theta:\mathfrak g\to V$, $\theta(X)=X\cdot 0$, is an isomorphism of vector spaces. Let $U=\theta(\mathfrak n)$ and $W=U^\perp$. If $U=0$, then $\mathfrak n=0$ and $d\pi[\mathfrak g]$ is an Abelian subalgebra of $\mathfrak{so}(V)$ of dimension $\dim\mathfrak g=\dim V$. However, the rank of $\mathfrak{so}(V)$ is at most $\frac12\dim V$, so this cannot be. It follows that $U\neq0$.

Set $\varphi=d\pi\circ\theta^{-1}$. Note that $\varphi$ is a homomorphism between the Abelian Lie algebras $V$ and $d\pi[\mathfrak g]$. It is also clear that $\mathfrak g$ has the required form. Of course, $\ker\varphi=\theta(\ker d\pi)=\theta(\mathfrak n)=U$.

Let $u\in U$, $w\in W$. Then $$[\tau_w+\varphi(w),\tau_u]=\tau_{\varphi(w)u} \in\mathfrak g, $$ so $\varphi(w)u\in \ker\varphi=U$. This shows that $\mathrm{im}\,\varphi$ preserves $U$ and, hence, the orthogonal decomposition $ V=U\oplus W$. Finally, if $w$, $w'\in W$, then $[\varphi(w),\varphi(w')]=0$, and so $$ [\tau_w+\varphi(w),\tau_{w'}+\varphi(w')]=\tau_{\varphi(w)w'} -\tau_{\varphi(w')w}=\tau_{\varphi(w)w'-\varphi(w')w} \in\mathfrak g. $$ This shows that $\varphi(w)w'-\varphi(w')w\in W\cap U=0$. Now the trilinear form on $W$ $$ (w,w',w'')\mapsto \langle\varphi(w)w',w''\rangle $$ is symmetric in the first two variables and skew-symmetric in the last two variables, hence, identically zero. This proves that $\varphi(W)W=0$ and completes the proof of the theorem. q.e.d.