 Solution

A Cartesian second-rank tensor is an isotropic tensor if its components are unchanged under every proper rotation matrix $R$:The Kronecker delta is isotropic because $R_{ik}{R}_{jk}={\delta }_{ij}$, the orthogonality relation $R{R}^T=I$.

No symmetry of $T$ needs to be assumed. A half-turn about the first axis has matrix $\mathrm{diag}(1,-1,-1)$; 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 $T=\mathrm{diag}(d_1,d_2,d_3)$. 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 isConversely, every member of this family is invariant because $R(\lambda I)R^T=\lambda I$.