Uniqueness of orthogonal decomposition of a symmetric matrix

Question

Let $A\in M_{n\times n}(\mathbb{R})$ be a symmetric and invertible matrix, and $\lambda_i, i = 1,..,n$ its eigenvalues. Let $D = \operatorname{diag}(\lambda_1,...,\lambda_n)$. We want to find matrices $X\in M_{n\times n}(\mathbb{R})$ such that $$XAX^T = D .$$ Since $A$ is symmetric, there exists a solution $X=P$ where $P$ is a matrix of normalized eigenvectors. For fixed $D$, when do we have uniqueness of the solution $X$? If the eigenvalues $\lambda_i$ can be repeated (i.e. $\lambda_i = \lambda_j$ for some $i\neq j$), we don't have uniqueness, for instance with $A=I$, $X$ can be any orthogonal matrix. However if $\lambda_i\neq \lambda_j, \forall i\neq j$, then is $X$ unique? If yes, is there a weaker condition to get uniqueness?

Answer 1

I think it is never unique. You define $P$ as the matrix of the normalized eigenvectors, but the eigenvectors can be chosen up to a sign. If you replace a column of $P$ by its opposite, it will still verify what you want.