= Solution
For $R=|x|>0$, $DR=\theta$ and $D\theta=(I-\theta\otimes\theta)/R$. Differentiating the <spherically symmetric deformation> gives
$$
\boxed{F=r'(R)\theta\otimes\theta+\frac{r(R)}R(I-\theta\otimes\theta).}
$$
For an orientation-preserving <spherically symmetric deformation> with positive radius, $r>0$ and $r'>0$. The radial direction and its perpendicular plane are <eigenspaces> of this positive <symmetric matrix>, so
$$
\boxed{\lambda_1=r'(R),\qquad\lambda_2=\lambda_3=r(R)/R.}
$$
At a positive <diagonal matrix> $D$, every diagonal <rotation> with two minus signs commutes with $D$. The supplied equivariance relation therefore forces $\widehat T(D)$ to be a <diagonal matrix>: each off-diagonal entry changes sign under one such <rotation> while $D$ stays fixed. Varying a positive diagonal entry shows that the diagonal entries of $\widehat T(D)$ are $\Phi_{,j}$. <Permutation> <symmetry> of $\Phi$ implies $\Phi_{,2}=\Phi_{,3}$ when $\lambda_2=\lambda_3$.
Choose a <rotation> whose first column is $\theta$. Applying equivariance to $F=QDQ^T$ now gives
$$
\boxed{T=\Phi_{,1}\theta\otimes\theta+\Phi_{,2}(I-\theta\otimes\theta),}
$$
with the <derivatives> evaluated at $(r',r/R,r/R)$. Thus the <stress> has one radial and one repeated tangential principal value, independently of the choice of tangent basis.
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