= Solution
Let $Q$ be the <orthogonal matrix> taking old components to new ones, so $x'_i=Q_{ip}x_p$. With repeated indices summed, the <tensor component transformation law> is
$$
\boxed{t'_{ij}=Q_{ip}Q_{jq}t_{pq},\qquad u'_{ijkl}=Q_{ip}Q_{jq}Q_{kr}Q_{ls}u_{pqrs}.}
$$
An <isotropic tensor> has the same components after every proper <rotation>. Since $Q_{ip}Q_{jp}=\delta_{ij}$, the <Kronecker delta> is invariant. Applying this identity to each factor shows that $a\delta_{ij}$ and every linear combination $b\delta_{ij}\delta_{kl}+c\delta_{ik}\delta_{jl}+d\delta_{il}\delta_{jk}$ are invariant as well.
To show that an <isotropic second-rank tensor> must be scalar, start with any matrix $t$, without assuming symmetry. Invariance under quarter-turns about axes 1 and 2 implies invariance under their squares, the half-turns
$$
Q_1^2=\operatorname{diag}(1,-1,-1),\qquad Q_2^2=\operatorname{diag}(-1,1,-1).
$$
The first forces $t_{12},t_{21},t_{13},t_{31}$ to vanish. The second also forces $t_{23},t_{32}$ to vanish. The matrix is therefore diagonal. A quarter-turn about axis 1 interchanges its 22 and 33 entries; one about axis 2 interchanges its 11 and 33 entries. All three must agree, giving $t=aI$.
A <rotation> preserves the ball $V_R$, $|x|^2$ and the <volume element>. Changing integration variables by that rotation shows that $A$ and $B$ are <isotropic tensors>. Write $A_{ij}=a\delta_{ij}$. Taking its <tensor contraction> gives
$$
3a=A_{ii}=\int_{V_R}|x|^4dV=4\pi\int_0^Rr^6dr=\frac{4\pi R^7}{7},
$$
so $a=4\pi R^7/21$. Every permutation of the indices of $B$ leaves its integrand unchanged. Its complete symmetry therefore makes all three isotropic coefficients equal, say $d$:
$$
B_{ijkl}=d(\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}).
$$
Double contraction gives $15d=B_{iikk}=\int_{V_R}|x|^4dV=4\pi R^7/7$. Thus the <isotropic fourth moment of a ball> and its radial contraction are
$$
\boxed{A_{ij}=\frac{4\pi R^7}{21}\delta_{ij},\qquad
B_{ijkl}=\frac{4\pi R^7}{105}(\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}).}
$$
Put $d=4\pi R^7/105$, so that $a=5d$. Contract $B$ with the prescribed <unit vector> twice:
$$
C_{ij}=A_{ij}+B_{ijkl}n_kn_l=d(6\delta_{ij}+2n_in_j).
$$
This has the <spectrum of a scalar matrix plus a rank-one projector>. In the direction $n$, the <outer product> $nn^T$ acts as the identity; on the plane perpendicular to $n$, it acts as zero. Therefore
$$
\boxed{Cn=\frac{32\pi R^7}{105}n,\qquad Cv=\frac{8\pi R^7}{35}v\quad\text{for }v\cdot n=0.}
$$
Any nonzero multiple of $n$ is a parallel <eigenvector>, and any nonzero vector in the perpendicular plane is a transverse <eigenvector>. The transverse <eigenvalue> has multiplicity two; an <orthonormal basis> of that plane completes a set of principal axes.
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