= Solution
Use the quarter-turn <rotation matrix>
$$
R=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix}.
$$
For a general second-order <tensor>, direct multiplication gives
$$
RTR^T=\begin{pmatrix}T_{22}&-T_{21}&-T_{23}\\-T_{12}&T_{11}&T_{13}\\-T_{32}&T_{31}&T_{33}\end{pmatrix}.
$$
Requiring $RTR^T=T$ gives $T_{11}=T_{22}$, $T_{12}=-T_{21}$, $T_{13}=-T_{23}$ together with $T_{23}=T_{13}$, and $T_{31}=-T_{32}$ together with $T_{32}=T_{31}$. Thus all mixed axial-transverse entries vanish and
$$
\boxed{T=\begin{pmatrix}\alpha&\omega&0\\-\omega&\alpha&0\\0&0&\beta\end{pmatrix}.}
$$
For an arbitrary rotation angle $\theta$, the transverse rotation is $Q_\theta=\cos\theta\,I+\sin\theta\,J$, where $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$. The transverse block of $T$ is $\alpha I-\omega J$, so it commutes with $Q_\theta$. Therefore $Q_\theta(\alpha I-\omega J)Q_\theta^T=\alpha I-\omega J$, while the axial component $\beta$ is unchanged.
Hence \b[invariance holds under every rotation about the axis], not just the quarter-turn. This is the general <axisymmetric second-order tensor>; a <symmetric tensor> would additionally require $\omega=0$, but that restriction was not assumed.
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