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Let $R_{ij}$ be the components of an orthogonal matrix relating two Cartesian frames by $v_i^{\prime }=R_{ij}{v}_j$. Repeated indices are summed from $1$ to $3$. The tensor component transformation law for a Cartesian second-rank tensor isFor 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 $S_x=\mathrm{diag}(1,-1,-1)$. Invariance under $S_{x}A{S}_x^{\mathsf{T}}$ 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 $A=\mathrm{diag}(a,b,c)$. 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