= Solution
Under a proper <rotation matrix> $R$, invariance of a <Cartesian second-rank tensor> reads $T=RTR^T$, equivalently $TR=RT$. The half-turn $R_z(\pi)=\operatorname{diag}(-1,-1,1)$ forces all entries mixing the $z$ direction with the $xy$ plane to vanish. Thus $T$ consists of a planar $2\times2$ block $B$ and the entry $T_{33}=c$.
Commutation with the planar quarter-turn $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ gives $B=aI+bJ$. This block already commutes with every planar rotation. Now use the half-turn $R_x(\pi)=\operatorname{diag}(1,-1,-1)$: on the planar block its conjugation fixes $I$ and sends $J$ to $-J$, so invariance forces $b=0$. Therefore
$$
\boxed{T=\operatorname{diag}(a,a,c),\qquad
t_{ij}=\alpha\delta_{ij}+\beta\delta_{i3}\delta_{j3},\quad
\alpha=a,\ \beta=c-a.}
$$
The formula uses the <Kronecker delta> and is sufficient as well as necessary: rotations preserving the unoriented $z$-axis fix both $I$ and $e_3e_3^T$. This is an <axially invariant second-rank tensor> with the planar antisymmetric part removed by the horizontal half-turn.
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