Suppose that matrix $A$ is diagonalizable, i.e. there exists an invertible matrix $P$ so that $P^{-1}AP$ is diagonal. We know that $P$ is not unique.
I am asking in which condition(s), $P$ is unique.
Here I propose a condition that diag($P$) = $(1, ..., 1)^T$ but it is not correct (according to one answer)
A one line answer: $P$ is NEVER unique. More precisely
let $A\in M_n(\mathbb{C})$ be diagonalizable and s.t. $spectrum(A)=(\lambda_i)$.
There is $P_0\in GL_n$ s.t. $P_0^{-1}AP_0=D=diag(\lambda_1,\cdots,\lambda_n)$.
Then the non-void set $Z=\{P\in GL_n;AP=PD\}$ is an algebraic set of dimension $\geq n$. In other words, the matrices that diagonalize $A$ in the form $D$ depend, at least, on $n$ complex parameters.
Moreover, $A$ can be diagonalized in the form $diag((\mu_i))$ where $\mu_i$ is a permutation of the $(\lambda_i)$; this gives many more solutions for $P$.
$\textbf{Remark}$. Of course, if you are only interested by the fact that $P$ is not unique, it suffices to remark that if $P$ works, then, for every $\alpha\not= 0$, $\alpha P$ works.