Let $V$ be a $d$-dimensional real vector space with inner product $\langle\cdot{,}\cdot\rangle$, and suppose $A$ is a symmetric linear map $V\to V$. Then, by the spectral theorem, $A$ has an orthonormal basis of eigenvectors.
Clearly, if $A$ is symmetric, so is $AA$. Suppose further that the eigenvalues $\lambda_{1} \dotsc,\lambda_{d}$ of $AA$ are all distinct, so that every eigenspace is one-dimensional. In other words, $AA$ has an orthonormal basis $(e_{1},\dotsc,e_{d})$ of eigenvectors, and such basis is unique (up to permutation and reflection).
Now, since every eigenvector of $A$ is an eigenvector of $AA$, we may conclude, by the uniqueness of $(e_{1},\dotsc,e_{d})$, that $e_{i}$ is an eigenvector of $A$ for every $i \in \{1,\dotsc,n\}$.
Question: Relaxing the hypothesis that the eigenvalues of $AA$ are all distinct, and assuming only that $\lambda_{j} \neq \lambda_{k}$, can we conclude that $e_{j}$ and $e_{k}$ are eigenvectors of $A$?
Suppose that $B$ is an $n\times n$ matrix with distinct eigenvalues. If $A$ is an $n\times n$ matrix which commutes with $B$ then $A$ preserves each eigenspace of $B$, so $A$ and $B$ can be simultaneously diagonalised. Indeed $A$ can be expressed as a polynomial in $B$, by say Lagrange interpolation.
In this example, take $B=A^2$. Then $AB=BA$ and so each eigenvector of $B$ is an eigenvector of $A$.
For your last question: consider $$A=\pmatrix{0&1&0\\1&0&0\\0&0&2}.$$ Then we can take $e_1,\ldots,e_3$ to be the standard basis, but $e_1$ is not an eigenvector of $A$.