Trace($A$) = Trace($A^2$) = Trace($A^3$)

Question

Suppose $A$ be an $n\times n$ matrix with real eigenvalues such that $\text{trace}(A)=\text{trace}(A^2)=\text{trace}(A^3)$. What can we conclude about the eigenvalues of A?

My attempts:

1. Considered $A$ as a diagonal matrix and $\lambda_1,\lambda_2 ,\ldots,\lambda_n$ to be its eigenvalues. So we get $$\lambda_1+\lambda_2+\cdots\lambda_n=\lambda_1^2+\lambda_2^2+\cdots+\lambda_n^2=\lambda_1^3+\lambda_2^3+\cdots+\lambda_n^3$$ From the first equality and using Cauchy–Schwarz inequality, we could get $\lambda_1+\lambda_2+\cdots\lambda_n\leq n$. My guess is that all non-zero eigenvalues are to be 1(just an intuition). But can't proceed any further.
2. Suppose $A$ is a $2\times 2$ matrix, then we could get trace($A^2$) = (trace($A))^2-2\det(A)$ and trace($A^3$) = (trace($A))^3-(3\det(A)\cdot\text{trace}(A))$. here also couldn't figure out a way.

Thank you in advance.