Let $$GL_{n}(\mathbb{R})=\{A_{n\times n}\mid A \text{ invertible matrix}\}$$ and $$O(n)=\{A\in GL_{n}(\mathbb{R})\mid A^{T}A=I_{n}\}$$
I am trying to make $O(n)$ as an $GL_{n}(\mathbb{R})$-module with the multiplication map $$f\colon GL_{n}(\mathbb{R})\times O(n)\to O(n)$$ defined by $f(A,B)=ABA^{-1}$ for every $A\in GL_{n}(\mathbb{R})$ and $B\in O(n)$.
First, I have to prove that the map is well-defined, i.e., $ABA^{-1}\in O(n)$.
Let $A\in GL_{n}(\mathbb{R})$ and $B\in O(n)$. We have to prove that $(ABA^{-1})^{T}(ABA^{-1})=I_{n}$.
From this, I get $$(ABA^{-1})^{T}(ABA^{-1})=(A^{-1})^{T}B^{T}A^{T}ABA^{-1}=(A^{T})^{-1}B^{T}A^{T}ABA^{-1}=(A^{T})^{-1}B^{T}(A^{T}A)BA^{-1}$$ But, $A$ just element in $GL_{n}(\mathbb{R})\not\in O(n)$. How to make $ABA^{-1}$ contained in $O(n)$? Is it possible to make $O(n)$ as an $GL_{n}(\mathbb{R})$-module?
Thank you in advance.