The correct answer is $\boxed{\text{A}}$.
The eigenvalues of a matrix are the roots of its characteristic polynomial. The characteristic polynomial of a matrix $A$ is given by $$p(x) = |xI – A|.$$
In this case, the characteristic polynomial of $S$ is $$p(x) = |xI – S| = \left| \begin{array}{cc} x – 3 & -2 \\ -2 & x – 3 \end{array} \right| = (x – 3)^2 – 4 = x^2 – 6x + 9.$$
The eigenvalues of $S$ are the roots of this polynomial, which are $5$ and $1$.
The eigenvalues of $S^2$ are the roots of the characteristic polynomial of $S^2$, which is $p(x)^2 = (x^2 – 6x + 9)^2 = x^4 – 12x^3 + 36x^2 – 108x + 81$.
The roots of this polynomial are $5$ and $1$, so the eigenvalues of $S^2$ are $\boxed{5}$ and $\boxed{1}$.
Here is a brief explanation of each option:
- Option A: The eigenvalues of $S^2$ are $5$ and $1$.
- Option B: The eigenvalues of $S^2$ are $6$ and $4$. This is not possible, because the sum of the eigenvalues of a matrix is equal to its trace, and the trace of $S^2$ is $6$.
- Option C: The eigenvalues of $S^2$ are $5$ and $1$. This is the correct answer.
- Option D: The eigenvalues of $S^2$ are $2$ and $10$. This is not possible, because the product of the eigenvalues of a matrix is equal to its determinant, and the determinant of $S^2$ is $9$.