= Solution
For a Cartesian change of components $x_i'=R_{ij}x_j$, a rank-$n$ <tensor> transforms according to
$$
\boxed{T'_{i_1\cdots i_n}=R_{i_1j_1}\cdots R_{i_nj_n}T_{j_1\cdots j_n},}
$$
with repeated indices summed. An <isotropic tensor> has unchanged components under every proper <rotation matrix>.
For rank two write $T'=RTR^T$. Invariance under quarter-turns implies invariance under their squares, the coordinate half-turns
$$
R_x(\pi)=\operatorname{diag}(1,-1,-1),\qquad R_y(\pi)=\operatorname{diag}(-1,1,-1).
$$
Each off-diagonal entry changes sign under at least one of these, so all off-diagonal entries vanish. A quarter-turn about the third axis then interchanges the first two diagonal entries, making them equal; a quarter-turn about the first axis equates the other two. Therefore
$$
\boxed{T_{ij}=\lambda\delta_{ij}.}
$$
Conversely $R(\lambda I)R^T=\lambda I$ for every rotation. This proves the <isotropic second-rank tensor> classification without assuming that the original <tensor> was symmetric. Also $\lambda=T_{kk}/3$ is a <scalar>, since the <trace> is rotation-invariant.
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