= Solution
For a Cartesian <change of basis> represented by an <orthogonal matrix> $Q$, <vectors> transform as $E'_i=Q_{ia}E_a$ and $B'_i=Q_{ia}B_a$. Their squared norms are invariant and $Q_{ia}Q_{jb}\delta_{ab}=\delta_{ij}$. Substitution into the given expression therefore gives
$$
\boxed{T'_{ij}=Q_{ia}Q_{jb}T_{ab},}
$$
the transformation law of a <Cartesian second-rank tensor>. The <magnetic field>'s extra axial sign under an improper physical reflection, if included, appears twice and cancels in its quadratic contribution.
Write the <Maxwell stress tensor> as $T=\mathbf E\mathbf E+\mathbf B\mathbf B-\tfrac12(E^2+B^2)I$. Its <divergence> is
$$
\mathbf M=(\mathbf E\cdot\nabla)\mathbf E+\mathbf E\nabla\cdot\mathbf E
+(\mathbf B\cdot\nabla)\mathbf B+\mathbf B\nabla\cdot\mathbf B
-\tfrac12\nabla(E^2+B^2).
$$
The identity $(\mathbf A\cdot\nabla)\mathbf A-\tfrac12\nabla A^2=-\mathbf A\times(\nabla\times\mathbf A)$ follows by contracting two <Levi-Civita symbols>, or directly by differentiating components. Using <Maxwell's equations> consequently yields
$$
\begin{aligned}
\mathbf M
&=\rho\mathbf E-\mathbf E\times(\nabla\times\mathbf E)-\mathbf B\times(\nabla\times\mathbf B)\\
&=\rho\mathbf E+\mathbf E\times\mathbf B_t-\mathbf B\times(\mathbf J+\mathbf E_t)\\
&=\rho\mathbf E+\mathbf J\times\mathbf B+\partial_t(\mathbf E\times\mathbf B).
\end{aligned}
$$
Hence the <local conservation of electromagnetic momentum> is
$$
\boxed{\partial_t(\mathbf E\times\mathbf B)=\mathbf M-\rho\mathbf E-\mathbf J\times\mathbf B,
\qquad M_i=\partial_jT_{ij}.}
$$
The two subtracted terms are the <Lorentz force density>, while $\mathbf E\times\mathbf B$ is the <electromagnetic momentum density> in the units used here.
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