The tensor operation shown in the red box is used in the textbook to prove that there are only 9 independent constants for orthotropic materials:

I want to use MMA to reproduce the operation of $C_{pqmn}=l_{ip}\;l_{jq}\;l_{km}\;l_{ln}\;C_{ijkl}$ (Where $C_{ijkl}$ is the stiffness tensor), but at present, I have no specific idea. I will continue to update the details to make it perfect.

Additional details:

Thanks to the help of xzczd, I updated some details of the problem. I want to know how to get 5 independent constants of transversely isotropic materials in the textbook when $l_{ij}$ is the rotation matrix of any angle below:

$$l_{ij}=\left( \begin{array}{ccc} \cos (\alpha ) & \sin (\alpha ) & 0 \\ -\sin (\alpha ) & \cos (\alpha ) & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$

Other details are being added...

The following formula is the constitutive relation of anisotropic materials:

$${\displaystyle {\begin{bmatrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _ {6}\end{bmatrix}}\,=\,{\begin {bmatrix} C_ {11}& C_ {12}& C_ {13}& C_ {14}& C_ {15}& C_ {16} \\ C_ {12}& C_ {22}& C_ {23}& C_ {24}& C_ {25}& C_ {26} \\ C_ {13}& C_ {23}& C_ {33}& C_ {34}& C_ {35}& C_ {36} \\ C_ {14}& C_ {24}& C_ {34}& C_ {44}& C_ {45}& C_ {46} \\ C_ {15}& C_ {25}& C_ {35}& C_ {45}& C_ {55}& C_ {56} \\ C_ {16}& C_ {26}& C_ {36}& C_ {46}& C_ {56}& C_ {66}\end {bmatrix}}{\begin{bmatrix}\varepsilon _{1}\\\varepsilon _{2}\\\varepsilon _{3}\\\varepsilon _{4}\\\varepsilon _{5}\\\varepsilon _ {6}\end{bmatrix}}}$$

It can be seen from the stiffness matrix that there are 21 independent elastic constants.

Orthotropic materials have three elastic symmetries which are perpendicular to each other(Orthotropic materials have three orthogonal planes of symmetry). In other words, if the coordinate axes are perpendicular to the three material symmetry planes of orthotropic materials, the elastic properties of the materials will not change after 180 degrees of rotation around these axes.

In the transformation process of the coordinate axis, the elastic symmetry of the material requires the fourth-order tensor to meet the following conditions:

$$C_{pqmn}=l_{ip}\;l_{jq}\;l_{km}\;l_{ln}\;C_{ijkl}$$

Where $l_{ij}$ is the symmetric transformation tensor.

For example, first of all, consider the material symmetry of orthotropic materials rotating 180 degrees around the Z axis, so $l_{ij}$ is :

$$l_{\text{ij}}=\left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$

From formula $C_{pqmn}=l_{ip}\;l_{jq}\;l_{km}\;l_{ln}\;C_{ijkl}$, we can get the following relations in this case:

$$C_{1311} = -C_{1311},\quad C_{1322} = -C_{1322} $$ $$C_ {1333} = -C_ {1333}, \quad C_ {1313} = -C_ {1313}$$ So, $$C_ {1311} = C_ {1322} = C_ {1333} = C_ {1313} = 0$$

the following relationships can also be obtained:

$$C_ {2311} = C_ {2322} = C_ {2333} = C_ {2312} = 0 \\ C_ {1213} = C_ {1223} = C_ {1123} = C_ {1113} = 0 \\ C_ {2223} = C_ {2213} = C_ {3323} = C_ {3313} = 0$$