= Solution
An <isotropic tensor> has unchanged components under every proper rotation of the Cartesian axes. Every rank-zero <scalar> is isotropic, and $c\delta_{ij}$ is a rank-two example. The zero rank-one tensor is isotropic, but no nonzero <vector> is: a half-turn about an axis perpendicular to it reverses its sign.
For the <inertia tensor>, write $u=x-G$ and $X=P-G$, so $x-P=u-X$. Expand its defining integrand. Every cross term has an integral containing $\int_Bu\rho\,d^3x=0$ by the definition of the <centre of mass>. The remaining terms give the <tensor parallel-axis theorem>:
$$
\boxed{I_{ij}(P)-I_{ij}(G)=M(|X|^2\delta_{ij}-X_iX_j).}
$$
For a uniform unit ball, rotational symmetry gives $\int\rho u_iu_jd^3u=c\delta_{ij}$. Its trace is $\int\rho|u|^2d^3u=3M/5$, so $c=M/5$ and $I(G)=2MI/5$. Therefore
$$
\boxed{I(P)=M\left[\left(\frac25+|X|^2\right)I-XX^T\right].}
$$
Assuming positive mass, it is isotropic exactly when \b[$P=G$]. For $X\ne0$, its <eigenvalues> and directions are
$$
\boxed{\lambda_\parallel=\frac{2M}{5},\quad e_\parallel=X/|X|;\qquad
\lambda_\perp=M\left(\frac25+|X|^2\right)\quad\text{on }X^\perp.}
$$
To give an explicit orthonormal <eigenbasis> when $X_1^2+X_2^2>0$, take $e_1=(-X_2,X_1,0)/\sqrt{X_1^2+X_2^2}$ and $e_2=e_\parallel\times e_1$. If $X$ is a nonzero <vector> along the third axis, take the first two coordinate <unit vectors> instead. At $X=0$, all three <eigenvalues> are $2M/5$ and every <orthonormal basis> works. The <inertia tensor of a displaced uniform ball> describes the same result for an arbitrary radius.
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