Let $H_1,H_2$ be arbitrary hyperplanes in $\mathbb{R}^n$ where $H_1 \cap H_2 \neq \{ \} $
Then, reflections $R_{H_1},R_{H_2}$ satisfies $R_{H_1}R_{H_2}=R(h,\alpha)$ where $h$ is some $n-2$ dimensional structure s.t $H_1 \cap H_2 = h$ and $R(h,\alpha)$ is a rotation about $h$ through angle $\alpha$.
Also, there exist another hyperplane $H_3$ s.t $R_{H_3}R_{H_1}=R(h,\alpha)$.
Thus, $R_{H_1}R_{H_2}=R_{H_3}R_{H_1}$.
This gives us $R_{H_1}R_{H_2}(R_{H_1})^{-1}=R_{H_3}$, so $R_{H_2}, R_{H_3}$ are conjugate.
And, as $H_1,H_2$ were chosen arbitrarily, any two reflections are conjugate.
This is just my rough idea, which needs more details. But, I couldn't really add any more details.
First of all, does my rough idea make sense?
Actually, my first goal was to show that $R_{H_1},R_{H_2}$ are conjugate, but the conclusion here is different from what I aimed for.
What do I need to consider if I want to show two arbitrary reflections are conjugate?
Any help would be appreciated!
Your approach is hard to understand. What is "some $n−2$ dimensional structure", for example?
But it looks like you are trying to prove with your approach that any two reflections about intersecting hyperplanes are conjugated by a reflection. That is true. Let $u, v$ be the normal vectors of the hyperplanes $H_u, H_w$. We can assume that $\|u\|=\|v\|=1$. Let $w$ be $u+v$ and $H_w$ be the hyperplane containing $H_u\cap H_v$ and perpendicular to $w$ (since $H_u\cap H_v$ is perpendicular to both $u$ and $v$, it is perpendicular to $w$). Then the reflection about $H_w$ will take $u$ to $v$, $H_u$ to $H_w$ and will conjugate the reflection about $H_u$ to the reflection about $H_v$.