I have second-order tensor defined in left-hand coordinate system. Now I need to use it in right-hand coordinate system. For simplicity, say it is Z-coordinate which is flipped.
For vectors (first-order tensor) it is obvious:
if
$$\overrightarrow {\mathbf V}_{LH}=[V_x, V_y, V_z]$$
then
$$\overrightarrow {\mathbf V}_{RH}=[V_x, V_y, -V_z]$$
How it will work for second-order tensor though? Assuming
$$ \mathbf T_{LH}= \begin{bmatrix} T_{xx} & T_{xy} & T_{xz} \\ T_{yx} & T_{yy} & T_{yz} \\ T_{zx} & T_{zy} & T_{zz} \\ \end{bmatrix} $$
then
$$ \mathbf T_{RH}= ? $$
It is know that any rank two tensor change as $B^{\top}TB$ for a base change $B$. In your case the matrix for $B$ is \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{bmatrix} So for $B^{\top}TB$ you are going to get
\begin{bmatrix} T_{xx} & T_{xy} & -T_{xz} \\ T_{yx} & T_{yy} & -T_{yz} \\ -T_{zx} & -T_{zy} & T_{zz} \\ \end{bmatrix}
You need to think that, in general, the components are related via $\vec V_{RH}=B^{-1}\vec V_{LH}$.