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Let $A$ be $n\times n$ positive-definite matrix over $\mathbb{R}$.

Does $\lbrace X\in M_{n\times n}(\mathbb{R})|X^TAX=A\rbrace\cong O(n;\mathbb{R})$ (homeomorphic)?

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Yes. If $Y$ is such that $A=Y^TY$ then $X\mapsto YXY^{-1}$ is an isomorphism of Lie groups from your stabilizer to $O(n)$.

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