= Solution
A <Cartesian second-rank tensor> is an <isotropic tensor> if its components are unchanged under every proper <rotation matrix> $R$:
$$
T_{ij}=R_{ik}R_{j\ell}T_{k\ell},\qquad\text{or equivalently }T=RTR^T.
$$
The <Kronecker delta> is isotropic because $R_{ik}R_{jk}=\delta_{ij}$, the orthogonality relation $RR^T=I$.
No symmetry of $T$ needs to be assumed. A half-turn about the first axis has matrix $\operatorname{diag}(1,-1,-1)$; invariance under it forces $T_{12},T_{13},T_{21},T_{31}$ to vanish. A half-turn about the second axis additionally kills $T_{23}$ and $T_{32}$. Thus $T=\operatorname{diag}(d_1,d_2,d_3)$. Each half-turn is the square of an allowed quarter-turn. A quarter-turn about the third axis interchanges $d_1,d_2$, so they are equal. A quarter-turn about the second axis interchanges $d_1,d_3$, so all three are equal. Consequently the complete family of <isotropic second-rank tensors> is
$$
\boxed{T_{ij}=\lambda\delta_{ij},\qquad\lambda\in\mathbb R.}
$$
Conversely, every member of this family is invariant because $R(\lambda I)R^T=\lambda I$.
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