= Solution
Choose an <orthonormal basis> with $\mathbf n=\mathbf e_3$. A half-turn about $\mathbf n$ kills all mixed axial-transverse components. Commutation with a quarter-turn makes the transverse block a scalar <identity matrix> plus an <antisymmetric matrix>. The resulting <oriented-axis invariant Cartesian tensor>, in coordinate-free form, is
$$
\boxed{\sigma_{ij}=a(\delta_{ij}-n_in_j)+d\,n_in_j+h\epsilon_{ikj}n_k,}
$$
with real constants $a,d,h$. The <Kronecker delta>, the <outer product> $n_in_j$, and the displayed <Levi-Civita symbol> contraction are invariant under every proper rotation about $\mathbf n$, so this form is also sufficient. In the chosen basis,
$$
\sigma=\begin{pmatrix}a&-h&0\\h&a&0\\0&0&d\end{pmatrix}.
$$
The antisymmetric part must be retained unless further symmetry is imposed.
Decompose the <electric field> as $\mathbf E=\mathbf E_\perp+\mathbf E_\parallel$, where $\mathbf E_\parallel=(\mathbf n\cdot\mathbf E)\mathbf n$. The <electric current density> is
$$
\mathbf J=a\mathbf E_\perp+d\mathbf E_\parallel+h\mathbf n\times\mathbf E_\perp.
$$
For a nonzero real <electric field>, the <alignment of current and field in an axisymmetric conductor> gives the complete classification of <collinearity>:
* If $\mathbf E_\perp=0$, then $\boxed{\mathbf J=d\mathbf E}$, with no additional symmetry required.
* If $\mathbf E_\parallel=0$, then $\boxed{\mathbf J\parallel\mathbf E\ \Longleftrightarrow\ h=0}$. Within this family $h=0$ is exactly the condition that the <conductivity tensor> is symmetric; then $\mathbf J=a\mathbf E$. An additional half-turn symmetry about a transverse axis is one way to force $h=0$.
* If both components are nonzero, then $\boxed{\mathbf J\parallel\mathbf E\ \Longleftrightarrow\ h=0\text{ and }a=d}$. These conditions give the <isotropic second-rank tensor> $\sigma=aI$, and then $\mathbf J=a\mathbf E$ for every direction.
For completeness, if $\mathbf E_\perp\ne0$ and $\mathbf J=\lambda\mathbf E$, taking the <dot product> with $\mathbf n\times\mathbf E_\perp$ forces $h=0$. The transverse equation then gives $\lambda=a$, while a nonzero axial component gives $\lambda=d$. This proves there are no missing directions. Zero current is included as the zero scalar multiple; if “parallel” requires a direction for $\mathbf J$, exclude cases where that scalar is zero. For a strictly anisotropic crystal the last possibility is absent.
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