eigenvectors of an eigenvector matrix Suppose I have a square matrix $A\in \mathbb{R}^{d\times d}$, with eigenvectors $v_1,v_2,\ldots,v_n$. Suppose I construct a new matrix $V = [v_1\ v_2\ \cdots\ v_n]$. Can anything be said about the eigenvalues or eigenvectors of this new matrix $V$. Do, $A$ and $V$ have same eigenvalues?
PS: I've not assumed A to be symmetric and $d$ can be greater than $n$. But if it helps derive something, please feel free to assume so or any other assumptions required.
PPS: I know that if $V$ is full rank, it diagonalizes $A$. I'm just wondering if there is any other relation.
Edit 1: Adding some special cases, that can be considered

*

*$A$ is real-symmetric, so that $v_1, v_2, \ldots$ are orthogonal and $d=n$.

*$A$ is normal (above holds).

 A: Not necessarily $A$ and $V$ will have same eigenvalues.
[ If $A$ is diagonalisable with an eigenvalue $0$ then one can construct $V$, invertible. Or, If $A$ is not diagonalisable with full rank then one can make $V$ with $0$ as an eigenvalue]
A: There is no reason for there to be any relation between $A$ and $V$ whatsoever. Of course $V$ is invertible since you are supposing the eigenvectors to form a basis (at least that is what I suppose, although the question does not actually say it), but apart from that it could be any matrix. And you can get loads of matrices satisfying the requirement that a given basis becomes one of eigenvectors: you can freely specify an eigenvalue independently for each of the vectors. This shows that the eigenvalues of $A$ cannot be determined even if $V$ is entirely known. Conversely, fixing $A$ you have quite some freedom for $V$, namely at least to independently scale each column, which completely messes up any eigenvalues that $V$ might have. If $A$ is a multiple of $I$, you can even choose $V$ to be any invertible matrix you want.
