Similar matrices C and D: how to derive the relation $\mathbf{x} = S^{-1} \mathbf{y}$ when $C = S^{-1}DS$

Question

D (with corresponding eigenvector $\mathbf{x}$) and C (with corresponding $\mathbf{y}$) are similar matrices, which means they have the same eigenvalues.

So the relation $C = S^{-1}DS$ holds.

So we get these relations: $C\mathbf{x} = \lambda \mathbf{x} $ and $ D\mathbf{y} = \lambda \mathbf{y}$

How can we with these details arrive to the equality?: $\mathbf{x} = S^{-1} \mathbf{y} $

Answer 1

Substituting $C$ with $D$ you get $ D S x = \lambda S x$, so $S x$ is an eigenvector of $D$.

Answer 2

An idea:

$$C=S^{-1}DS\iff CS^{-1}=S^{-1}D\implies \color{red}{CS^{-1}x}=S^{-1}Dx=S^{-1}(\lambda x)=\color{red}{\lambda S^{-1}x}$$

Thus, $\;S^{-1}x\;$ is an eigenvalue of $\;C\;$ corresponding to $\;\lambda\;$ . Now, if the corresponding eigenspaces (for $\;C, D\;$) are one-dimensional, then we already get $\;ky=S^{-1}x\;,\;\;k\in\Bbb F\;$, so I'm not sure the claim is true as written.