Statement I is false, since row operations do not preserve eigenvalues in general.
For example,
$M=\begin{pmatrix}1&0\\0&1\end{pmatrix}$
has eigenvalues $1,1$.
After interchanging the two rows, we get
$M'=\begin{pmatrix}0&1\\1&0\end{pmatrix}$,
whose eigenvalues are $1,-1$.
So Statement I is false.
Statement II is also false, since row operations on $(M-\lambda I)$ do not preserve eigenvalues in general.
Take
$M=\begin{pmatrix}1&0\\0&2\end{pmatrix}$ and $\lambda=1$.
Then
$M-\lambda I=\begin{pmatrix}0&0\\0&1\end{pmatrix}$,
whose eigenvalues are $0,1$.
After interchanging the two rows, we get
$(M-\lambda I)'=\begin{pmatrix}0&1\\0&0\end{pmatrix}$,
whose eigenvalues are $0,0$.
So Statement II is false.
Hence, the correct answer is $\boxed{\text{D. None}}$.