= Solution
An order-$n$ Cartesian <tensor> has $3^n$ components whose values in two orthonormal Cartesian frames satisfy
$$
T'_{i_1\ldots i_n}=R_{i_1j_1}\cdots R_{i_nj_n}T_{j_1\ldots j_n},\qquad R^TR=I,
$$
with repeated indices summed as in <Einstein notation>. It is an <isotropic tensor> when its components are unchanged by every proper <rotation>, $R\in SO(3)$.
For rank two, a half-turn about any coordinate axis changes the sign of each component having exactly one index along that axis. Invariance therefore makes every off-diagonal component zero. Quarter-turns interchanging two coordinate axes then require $T_{11}=T_{22}=T_{33}$. Thus, without assuming symmetry in advance,
$$
\boxed{T_{ij}=a\delta_{ij}}.
$$
Conversely, the <Kronecker delta> is invariant because $R_{ik}R_{jk}=\delta_{ij}$, so every tensor of this form is isotropic.
For rank three the general proper-rotation invariant form is
$$
\boxed{T_{ijk}=b\epsilon_{ijk}},
$$
where $\epsilon_{ijk}$ is the <Levi-Civita symbol>. The convention matters: the <isotropic rank-three tensor and reflections> distinction says that invariance of a polar <tensor> under all $R\in O(3)$ would force $b=0$. A nonzero $\epsilon_{ijk}$ is invariant under proper <rotations>, and is a <pseudotensor> under the full orthogonal group. The shell integrals below are invariant under reflections as well, so their odd moments vanish.
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