Let $f : \mathbb{C} \to \mathbb{C}$ linear map such $$ f(e_{i}) = \begin{cases} e_{i+1} & 1 \leq i<n \\ e_{1} & i=n\end{cases}$$ Diagonalize $f$.
Thoughts I know the characteristic polynomial is $x^n-1$ and the eigenvalues are of the form $e^{\frac{2i\pi k}{n}}\quad \forall k\in [1,n] $. So for $P^{-1}[f]_\epsilon P=D$ ($\epsilon$ in standard basis of $C$) I have the $D$ matrix, but how can I find $P$?
I assume that $e_1,e_2,\dots,e_n$ refers to the standard basis of $\Bbb C^n$.
Hint: Given $x = (x_1,\dots,x_n)$, the equation $f(x) = \lambda x$ becomes $$ (x_n,x_1,\dots,x_{n-1}) = \lambda(x_1,x_2,\dots,x_n) \implies \begin{cases} x_1 = \lambda x_2\\ \;\;\quad \vdots \\ x_{n-1} = \lambda x_n\\ x_n = \lambda x_1. \end{cases} $$ Because eigenvectors can be scaled arbitrarily, we can assume without loss of generality that $x_1 = 1$. You might find this system easier to solve if you divide both sides by $\lambda$ first, noting that $1/e^{2 \pi i k} = e^{-2 \pi i k}$.