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

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

1. Use $(ast)$ (twice) to show deduce that $A^4 = A$. Conclude that any eigenvalue of $A$ is a root of $$p(x) := x^4 - x = x (x - 1) (x^2 + x + 1) .$$
2. 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 $3$.)
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, $alpha^2$.
4. 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.

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.