= Solution
For <Cartesian tensors>, let $R$ be the <orthogonal matrix> taking old vector components to new ones, $v'_i=R_{ij}v_j$. A rank-$n$ <tensor> transforms by one copy of $R$ on each index:
$$
\boxed{T'_{i_1\cdots i_n}=R_{i_1j_1}\cdots R_{i_nj_n}T_{j_1\cdots j_n}.}
$$
Repeated indices are summed according to <Einstein notation>. In general coordinates a <contravariant tensor> index instead carries the corresponding <Jacobian matrix> $\partial x'^i/\partial x^j$, and a <covariant tensor> index carries its inverse; in <orthonormal bases> of Cartesian axes their component laws coincide with the displayed form.
An <isotropic tensor> has identical components after every proper <orthogonal transformation>, so $T'=T$ for every $R$ with $R^TR=I$ and $\det R=1$. Requiring reflection invariance as well is a stronger convention for odd tensor ranks. The <Kronecker delta> is invariant under all <orthogonal transformations>, while the <Levi-Civita symbol> is invariant under proper rotations and acquires a determinant sign under reflections.
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