= Solution
Let $R_{ij}$ be the components of an <orthogonal matrix> relating two Cartesian frames by $v'_i=R_{ij}v_j$. Repeated indices are summed from $1$ to $3$. The <tensor component transformation law> for a <Cartesian second-rank tensor> is
$$
\boxed{A'_{ij}=R_{ik}R_{j\ell}A_{k\ell}},\qquad A'=RAR^{\mathsf T}.
$$
For a proper <rotation>, $\det R=1$; the transformation law itself also holds for orthogonal changes of Cartesian frame with determinant $-1$.
To find a <cubically invariant second-rank tensor>, first square each of the prescribed quarter-turns. For example the half-turn about the first axis is $S_x=\operatorname{diag}(1,-1,-1)$. Invariance under $S_xAS_x^{\mathsf T}$ forces $A_{12},A_{13},A_{21},A_{31}$ to vanish. The corresponding half-turns about the other axes eliminate the remaining off-diagonal components, even though symmetry of $A$ was never assumed.
Now write $A=\operatorname{diag}(a,b,c)$. A quarter-turn about the third axis interchanges $a,b$, so $a=b$; a quarter-turn about the first axis gives $b=c$. Conversely a scalar multiple of the <identity matrix> is invariant under every <rotation>. Thus the most general answer is
$$
\boxed{A_{ij}=a\delta_{ij}}.
$$
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