= Solution
A Cartesian <isotropic tensor> is unchanged by every proper <rotation matrix>. For a rank-two tensor this means $B'=RBR^{\mathsf T}=B$, or $B'_{ij}=R_{ik}R_{jl}B_{kl}=B_{ij}$. No symmetry of $B$ is assumed. For a quarter-turn about the $z$ axis,
$$
R_z=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix},\qquad
R_zBR_z^{\mathsf T}=\begin{pmatrix}
B_{22}&-B_{21}&-B_{23}\\
-B_{12}&B_{11}&B_{13}\\
-B_{32}&B_{31}&B_{33}
\end{pmatrix}.
$$
Comparing entries yields $B_{11}=B_{22}$, $B_{12}=-B_{21}$, $B_{13}=-B_{23}= -B_{13}$, and $B_{31}=-B_{32}=-B_{31}$. In particular,
$$
\boxed{B_{13}=B_{31}=B_{23}=B_{32}=0}.
$$
The remaining matrix has the form $\begin{pmatrix}b&c&0\\-c&b&0\\0&0&d\end{pmatrix}$. A quarter-turn about the $x$ axis gives
$$
R_x=\begin{pmatrix}1&0&0\\0&0&-1\\0&1&0\end{pmatrix},\qquad
R_xBR_x^{\mathsf T}=\begin{pmatrix}b&0&c\\0&d&0\\-c&0&b\end{pmatrix}.
$$
Invariance now forces $c=0$ and $d=b$, proving that an <isotropic second-rank tensor> is
$$
\boxed{B_{ij}=b\delta_{ij}},
$$
a scalar multiple of the <Kronecker delta>.
For the final unheaded conductivity identity, let $P=nn^{\mathsf T}$, the orthogonal <projection> onto the unit vector $n$. Then $P^2=P$ and the <uniaxial conductivity tensor> is $\sigma=\alpha I+\gamma P$. The antisymmetric tensor $D$ sends a vector $v$ to $v\times n$. Applying it twice and using the vector triple product gives
$$
D^2v=(v\times n)\times n=n(n\cdot v)-v,
\qquad D^2=P-I.
$$
Consequently
$$
\sigma D^2=(\alpha I+\gamma P)(P-I)=\alpha P-\alpha I.
$$
This equals $-\sigma=-\alpha I-\gamma P$ exactly when $(\alpha+\gamma)P=0$. Since $P$ has rank one and is nonzero, the requested parameter is
$$
\boxed{\gamma=-\alpha}.
$$
At this value conductivity vanishes along $n$; it is precisely the singular case excluded in part (ii), so there is no conflict with that part's conclusion.
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