= Solution
Identify $\mathbb R^4$ with the <quaternions>, and use the identification of <unit quaternions> with $SU(2)$ proved in the root Solution. The action
$$
(q_L,q_R):v\longmapsto q_Lvq_R^{-1}
$$
preserves the Euclidean <norm> by multiplicativity of <quaternion> <norms>. It is a homomorphism $SU(2)\times SU(2)\to SO(4)$: its domain is connected, so the orthogonal image has <determinant> $+1$.
An element in the kernel fixes $v=1$, giving $q_L=q_R=q$. Fixing every other $v$ means $q$ commutes with all <quaternions>, so $q$ is real; since it has unit <norm>, $q=\pm1$. Hence the kernel is the diagonal subgroup $\{(1,1),(-1,-1)\}$.
The differential acts as $v\mapsto av-vb$ for imaginary <quaternions> $a,b$. If it vanishes, $v=1$ gives $a=b$, and commutation with all $v$ makes $a$ real and imaginary, hence zero. The differential is injective, and both <Lie algebras> have dimension six. The image is therefore an open subgroup of connected $SO(4)$, hence all of $SO(4)$. Thus
$$
\boxed{SO(4)\cong\bigl(SU(2)\times SU(2)\bigr)/\mathbb Z_2,}
$$
where $\mathbb Z_2$ acts diagonally, not separately on the two factors.
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