Let $V$ be $n$ dimensional vector space, with basis $v_1,\cdots,v_n$. Suppose $\mathscr{A}: V\to V$ is an invertible linear operator, such that $\mathscr{A}(v_i)\in \{v_1,\cdots,v_n\}$, $i=1,\cdots,n$. Show that $\mathscr{A}$ is diagonal, and has roots of $1$ as eigenvalues.
My attempt: show that $v_1,\mathscr{A}v_1,\cdots,\mathscr{A}^{n-1}v_1$ are linearly independent, and then has eigenvalues: $n$ roots of $1$, so $\mathscr{A}$ can be diagonalisable. But how?
This proof works on the assumption that your question was defined over $\mathbb{C}$. Because $\mathscr{A}$ is invertible, it must map bases to bases, and given the further restriction that $\mathscr{A} v_i=v_j$ for some $1 \leq i,j \leq n$, we can find a permutation $\pi$ such that $\mathscr{A} v_i=v_{\pi{(i)}}$. Then, in the basis $\{v_i, \ldots, v_n\}$, $\mathscr{A}$ can be represented as the permutation matrix:
$$P_{\pi}= \begin{bmatrix} \mathbb{e_{\pi{(1)}}}\\ \vdots\\ \mathbb{e_{\pi{(n)}}}\end{bmatrix}$$
This matrix has orthogonal columns, and thus acts as an isometry, i.e., $\|P_{\pi} x\|=\|x\|\ \ \forall \ x$. If $x$ were an eigenvector of $P_{\pi}$, we would have $\|P_{\pi} x\|=\|\lambda x\| \implies |\lambda|=1$. Depending on the order $m$ of the permutation, your eigen values would be $e^{2 \pi i k/m}$.