= Solution
Use the right-handed <rotation matrix> which sends the positive $x$ direction to the positive $y$ direction:
$$
R=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix}.
$$
The <Cartesian second-rank tensor> transformation law gives
$$
S'=RSR^{\mathsf T}=\begin{pmatrix}
S_{22}&-S_{21}&-S_{23}\\
-S_{12}&S_{11}&S_{13}\\
-S_{32}&S_{31}&S_{33}
\end{pmatrix}.
$$
Here <isotropic tensor> means invariant under every proper <rotation>, as is needed for the rank-three assertion involving the <Levi-Civita symbol>. A passive coordinate <rotation> through the same angle uses $R^{\mathsf T}$ instead; the invariance conclusions are identical.
Setting $S'=S$ forces $S_{11}=S_{22}$, $S_{12}=-S_{21}$, and $S_{13}=S_{23}=S_{31}=S_{32}=0$. No symmetry assumption on $S$ has been made: it now has the form
$$
S=\begin{pmatrix}a&b&0\\-b&a&0\\0&0&c\end{pmatrix}.
$$
Invariance under a right-handed quarter-turn about the $x$ axis next gives $S_{22}=S_{33}$ and $S_{12}=-S_{13}$, hence $a=c$ and $b=0$. Conversely $R(\lambda I)R^{\mathsf T}=\lambda I$ for every <orthogonal matrix>. The general <isotropic second-rank tensor> is therefore \b[$S_{ij}=\lambda\delta_{ij}$].
An isotropic <vector> must be fixed by the half-turn about the $z$ axis, so its first two components vanish. The half-turn about the $x$ axis then forces its third component to vanish. Thus \b[the only isotropic <vector> is zero]. The general <isotropic third-rank tensor> under proper <rotations> is \b[$A_{ijk}=\lambda\epsilon_{ijk}$]. If invariance under the whole <orthogonal group> were required of an ordinary rank-three <tensor>, inversion $R=-I$ would send $A$ to $-A$, forcing $A=0$ instead.
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