This was [asked](https://math.stackexchange.com/questions/3012437/eigenvectors-of-nth-root-of-identity-matrix) at math stackexchange a long time ago with no answers but some upvotes.


Let $A^n=I,$ where $A$ is $n\times n,$ and assume that $A^k\neq I,$ for all $1\leq k<n.$ Since its characteristic polynomial is $x^n-1$, the distinct $n^{th}$ roots of unity are its eigenvalues, 


Thus there are a full set of linearly independent eigenvectors.


* What do they look like?
* If we assume $A$ is orthogonal, what do they look like?
* If we assume $A$ is real, can one say anything more?



What do the eigenvectors of the $n$th roots of $I_n$ look like?
Since $A^n=I$, $A$ is diagonalizable and eigenvalues are $n$-th roots of unity. They don't have to be all $n$-th roots of unity and they don't have to be distinct.
Your condition that $A^k\neq I$ simply means that for all $k<n$ not all
$\lambda\_j^k$ are equal to $1$.


Speaking on eigenvectors, there are no restrictions: let $D$ be a diagonal matrix
whose elements are $n$-th roots of unity, and $B$ an arbitrary non-singular matrix.
Then $A=BDB^{-1}$ has the property that $A^n=I$ and eigenvectors are columns of $B$.


If $A$ is real, we have an additional property that eigenvalues come in conjugate pairs, and to conjugate eigenvalues correspond conjugate eigenvectors. Any $D$ and $B$ with these properties will give you $A$ with the required properties.


If $A$ is real and orthogonal, then we have the additional property that eigenvectors can be chosen to form an orthonormal basis. This means that $B$ above can be chosen unitary, so $A$ us unitary, and as it is real, it is orthogonal.


This is a complete description of eigenvalues and eigenvectors.


