Answer
See below
Work Step by Step
Let $v$ be an eigenvector of A corresponding to the eigenvalue $\lambda$.
We have $B=S^{-1}AS$
then $\rightarrow B(S^{-1}v)=(S^{-1}AS)(S^{-1}v)\\
\rightarrow B(S^{-1}v)=S^{-1}(Av)\\
\rightarrow B(S^{-1}v)=S^{-1}(\lambda v)=1\\
\rightarrow B(S^{-1}v)=\lambda(S^{-1}v)$
Hence, $S^{-1}v$ is an eigenvector of $B$ corresponding to the eigenvalue $\lambda$
