If $A$ is a matrix such that $A^T = A^2$, what are eigenvalues of $A$? 
If $A$ is a matrix such that $A^T = A^2$, what are eigenvalues of $A$?

Now I read somewhere that changing the matrix by taking a transpose does not change the characteristic polynomial. So it is safe to say that the annihilating polynomial in this case is
$x^2 - x$.
If this is the case then I think the answer is quite easy. But if there is some caveat I am missing then I am lost as to how to approach this problem.
 A: It is indeed exercise 4.4.14 in Artin's "Algebra", US edition.
If only possible eigenvalues are of interest then from $A^T=A^2$ we can deduce that $A^4=A$ (applying the transpose to both sides). Hence all eigenvalues are roots of $t^4-t=t(t^3-1)$. So $0, \sqrt[3]{1}$ (all three roots of 1) are the possible eigenvalues, and all 4 of these can occur.
But one can deduce more. From $A^T=A^2$ one can see that $A$ and $A^T$ commute, so $A$ is a normal matrix, hence it is unitarily similar to a diagonal matrix.
A: We cannot conclude that the polynomial $x^2 - x$ annihilates $A$, as Angina Seng's example
$$P := \pmatrix{\cdot&1&\cdot\\\cdot&\cdot&1\\1&\cdot&\cdot}$$
in the comments shows: It's true that $A$ and $A^\top$ have the same minimal polynomials and so are annihilated by the same polynomials, but this does not allow us to replace $A^\top$ with $A$ in the given condition $$\phantom{(ast)} \qquad A^\top = A^2 . \qquad (\ast)$$
Hint

*

*Use $(\ast)$ (twice) to show deduce that $A^4 = A$. Conclude that any eigenvalue of $A$ is a root of $$p(x) := x^4 - x = x (x - 1) (x^2 + x + 1) .$$

*Denote a root of $x^2 + x + 1$ by $\alpha$; verify that $\alpha^2 = -\alpha - 1$ is also a root of $x^2 + x + 1$. In particular, the spectrum (set of eigenvalues) $\sigma(A)$ of $A$ is some subset $$\sigma(A) \subseteq \{0, 1, \alpha, \alpha^2\} .$$ (If this set has fewer than four elements, then necessarily $\alpha = \alpha^2 = 1$, and substituting establishes that this happens iff the underlying field has characteristic $3$.)

*Use $(\ast)$ again to show that if $\lambda \in \sigma(A)$ then $\lambda^2 \in \sigma(A)$, that is, $\sigma(A)$ is closed under the operation $\lambda \mapsto \lambda^2$. In particular, if one root $\alpha$ of $x^2 + x + 1$ is an eigenvalue of $A$, so is the other root, $\alpha^2$.

*Steps (1)-(3) imply that $\sigma(A)$ must be one of $7$ particular subsets of $\{0, 1, \alpha, \alpha^2\}$ (or, over characteristic $3$, one of the three nonempty subsets of $\{0, 1\}$). Angina Seng's example $P$ can be used to establish a few of the harder cases (over any field).

Remark Without more restrictions, there is a subtlety to (4): Which subsets of $\{0, 1, \alpha, \alpha^2\}$ can be realized as $\sigma(A)$ for some $A \in M(n, \Bbb F)$ satisfying $(\ast)$ can depend on the base field $\Bbb F$ and size $n \times n$ of $A$ (even beyond the evident restriction $|\sigma(A)| \leq n$).
For example, one can show that if $A \in M(2, \Bbb R)$ satisfies $(\ast)$ and $\alpha \in \sigma(A)$, then $A$ is a (real) rotation matrix. There are two such matrices that satisfy $(\ast)$, but both have some irrational entries, so there is no $A \in M(2, \Bbb Q)$ satisfying $(\ast)$ with $\alpha \in \sigma(A)$. On the other hand, for any $\Bbb F$ (with $\operatorname{char} \Bbb F \neq 3$) and any $n \geq 4$, at least $5$ of the $7$ possibilities occur.
