If you are looking for validity of the equation for elastic tensor linear transformation, here is a quick and easy explanation:
From Hooke's law, we know that: $\sigma = \mathbf{C} \cdot \varepsilon$, where $\sigma$ is the stress tensor, $\varepsilon$ is the strain tensor, and $\mathbf{C}$ is the elastic tensor. Let's say you want to transform the coordinate linearly as:
$$\sigma^{'} = K_{\sigma} \sigma$$
$$\varepsilon^{'} = K_{\varepsilon} \varepsilon$$
Where $\sigma^{'}$ and $\varepsilon^{'}$ are stress and strain tensors in the new coordinate and $K_{\sigma}$ and $K_{\varepsilon}$ are the transformation matrices for stress and strain tensor respectively. We know that transforming the coordinates linearly should not have any effect on Hooke's law:
$$\sigma^{'} = \mathbf{C}^{*} \cdot \varepsilon^{'}$$
or:
$$K_{\sigma} \sigma = \mathbf{C}^{*} \cdot K_{\varepsilon} \varepsilon$$
or:
$$\sigma = K_{\sigma}^{-1} \mathbf{C}^{*} \cdot K_{\varepsilon} \varepsilon$$
But we had:
$$\sigma = \mathbf{C} \cdot \varepsilon$$
So:
$$\mathbf{C} = K_{\sigma}^{-1} \mathbf{C}^{*} K_{\varepsilon}$$
or:
$$\mathbf{C}^{*} = K_{\sigma} \mathbf{C} K_{\varepsilon}^{-1}$$
But $K_{\sigma}$ and $K_{\varepsilon}$ are connected together based on Reuter's matrices:
$$K_{\varepsilon} = R K_{\sigma} R^{-1}$$
Where $R$ and $R^{-1}$ are Reuter's matrices:
$$R = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 2
\end{bmatrix}$$
$$R^{-1} = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & \frac{1}{2}
\end{bmatrix}$$
So finally, you would have:
$$\mathbf{C}^{*} = K_{\sigma} \mathbf{C} R K^{-1}_{\sigma} R^{-1}$$
Check the validity of the derivation from Gerdeen's book at eq. 8.65.
But if you are looking to understand what will remain constant after linear transformation, I refer you to these blogs and Wikipedia page to find out tensor invariants are directly related to their eigenvalues: https://imechanica.org/node/6771 and https://en.wikipedia.org/wiki/Invariants_of_tensors
