To prove this we start with the eigenvalue problem and a similarity transformed matrix $\mathbf{B}$. $$\mathbf{A}\mathbf{x}=\lambda\mathbf{x} \hspace{1cm} \mathrm{and}\hspace{1cm} \mathbf{B}= \mathbf{S}^T \mathbf{A}\mathbf{S}.$$ We multiply the first equation on the left by $\mathbf{S}^T$ and insert $\mathbf{S}^{T}\mathbf{S} = \mathbf{I}$ between $\mathbf{A}$ and $\mathbf{x}$. Then we get $$\begin{equation} (\mathbf{S}^T\mathbf{A}\mathbf{S})(\mathbf{S}^T\mathbf{x})=\lambda\mathbf{S}^T\mathbf{x} , \tag{1} \end{equation}$$ which is the same as $$\mathbf{B} \left ( \mathbf{S}^T\mathbf{x} \right ) = \lambda \left (\mathbf{S}^T\mathbf{x}\right ).$$ The variable $\lambda$ is an eigenvalue of $\mathbf{B}$ as well, but with eigenvector $\mathbf{S}^T\mathbf{x}$.