If $\lambda$ is eigenvalue of matrix $A$ and $B=S^{-1} A S$ show that $\lambda$ is also eigenvalue of matrix $B$. Also show that if matrix $A^{2015}$ is a zero matrix then its only eigenvalue is $0$.

The first problem I tackled like this. if $\lambda$ is a eigenvalue of $B$ it solves this equation.
\begin{align\*}
B&=S^{-1} A S\;\;\;|\cdot x\\
Bx&=(S^{-1} A S) x\;\;\;\text{where }(S^{-1} A S)=\lambda
\end{align\*}
So this means that $\lambda=S^{-1} A S$ and it should solve:
\begin{align\*}
Ax&=(S^{-1} A S)x\\
A&=S^{-1}AS=B
\end{align\*}
And shouldn't this hold because now while writing that I was thinking that for all matrises $X$ if you multiply it first with matrix $S$ and then with $S^{-1}$ should it be $X$ so does this hold for all matrises?:

$$B=S^{-1}AS=A$$

Isn't the second question very trivial because $A^{2015}=\mathbf{0}$
\begin{align\*}
A^{2015}x&=\lambda x \\
\mathbf{0}x &= \lambda x \\
\mathbf{0} &= \lambda x\;\;\text{where } x\neq\mathbf{0} \text{ thus}\\
\lambda &= 0
\end{align\*}