= Solution
The radial gradient $F=D\mathbf y$ is symmetric positive definite. If it belongs to a well $SO(3)U_i$, write $F=QU_i$. Then $F^TF=U_i^2$, while positive symmetry makes $F$ itself the unique <positive square root of an operator> of $F^TF$. Hence $F=U_i$. Likewise a <symmetric positive-definite matrix> in $SO(3)$ must be $I$. Therefore
$$
D\mathbf y(\mathbf x)\in\{I,U_1,\ldots,U_N\}.
$$
For each radius, the tangential stretch $r(R)/R$ must belong to the finite set of eigenvalues of these matrices. The ratio is <continuous> on the <connected> radial interval, so it is one constant $v>0$. Thus $r(R)=vR$ and differentiation gives $r'=v$ as well. Hence
$$
\boxed{\mathbf y(\mathbf x)=v\mathbf x.}
$$
This proof uses finiteness and positive definiteness of the wells, not isotropy or the <Baker-Ericksen inequalities>. It also covers $R_1=0$: the punctured three-dimensional ball remains <connected>, and no condition at the excluded center was used.
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