= Solution
Let $\overline{\mathbf r}=a^2\mathbf r/r^2$ and $\widehat{\mathbf r}=\mathbf r/r$. Its Jacobian is
$$
\frac{\partial\overline r_i}{\partial r_j}
=\frac{a^2}{r^2}(\delta_{ij}-2\widehat r_i\widehat r_j).
$$
The <chain rule> therefore yields
$$
\nabla_{\mathbf r}
=\frac{a^2}{r^2}
\left[\nabla_{\overline{\mathbf r}}
-2\widehat{\mathbf r}(\widehat{\mathbf r}\cdot\nabla_{\overline{\mathbf r}})\right].
$$
Taking its <cross product> with $\mathbf r$ kills the radial term and proves that <spherical inversion preserves angular derivatives>:
$$
\boxed{\mathbf r\times\nabla_{\mathbf r}
=\frac{a^2}{r^2}\mathbf r\times\nabla_{\overline{\mathbf r}}
=\overline{\mathbf r}\times\nabla_{\overline{\mathbf r}}.}
$$
This is also the invariance of the differential expression underlying the <angular momentum operator>.
A scalar <spherical harmonic> $Y$ depends only on direction, so $Y(\widehat{\mathbf r})=Y(\widehat{\overline{\mathbf r}})$ and its gradient is tangential. Thus $\widehat{\mathbf r}\cdot\nabla_{\overline{\mathbf r}}Y=0$, and
$$
r\nabla_{\mathbf r}Y=\overline r\nabla_{\overline{\mathbf r}}Y,
\qquad \widehat{\mathbf r}=\widehat{\overline{\mathbf r}}.
$$
Each of the three normalized <vector spherical harmonic> families is consequently unchanged as an angular basis function:
$$
\boxed{\mathbf P(\overline{\mathbf r})=\mathbf P(\mathbf r),
\qquad \mathbf B(\overline{\mathbf r})=\mathbf B(\mathbf r),
\qquad \mathbf C(\overline{\mathbf r})=\mathbf C(\mathbf r).}
$$
The normalization requires $n\geq1$ for $\mathbf B$ and $\mathbf C$; at $n=0$ the gradient vanishes and only $\mathbf P$ remains.
This substitution into the basis functions is distinct from the <pushforward of a vector field> of a physical vector field by inversion. Under that Jacobian map the radial vector picks up $-a^2/r^2$ and the two tangential vectors pick up $+a^2/r^2$. Likewise, the weighted <Kelvin transform> of a scalar includes the extra factor $a/r$. Neither operation changes the angular differential identity just proved.
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