I have a vector A with direction $\vec D$.
I also have another vector B with direction $\vec R$.
(think of $\vec R$ to be the reflection of $\vec D$ at point B.)
From the reflection formula:$$ \vec R = \vec D-2*(\vec D \cdot \vec N)*\vec N $$ (we know both $\vec R$ and $\vec D$)
How can I find $\vec N$?
The reflection formula says that to reflect a vector $\mathbf v$ in the hyperplane specified by its normal $\mathbf n$, you reverse the component of $\mathbf v$ that’s in the direction of $\mathbf n$. That is, you decompose $\mathbf v$ into $\pi_{\mathbf n}\mathbf v+(\mathbf v-\pi_{\mathbf n}\mathbf v)$, where $\pi_{\mathbf n}$ is orthogonal projection onto $\mathbf n$, and then negate the first component: $$\mathbf v'=-\pi_{\mathbf n}\mathbf v+(\mathbf v-\pi_{\mathbf n}\mathbf v)=\mathbf v-2\pi_{\mathbf n}\mathbf v.$$ To recover the normal given $\mathbf v$ and $\mathbf v'$, remember that $\pi_{\mathbf n}\mathbf v$ is a scalar multiple of $\mathbf n$, so $\mathbf v-\mathbf v'=2\pi_{\mathbf n}\mathbf v$ is a multiple of $\mathbf n$ and is also normal to the hyperplane.
Without more information about $\mathbf n$ and its relationship to $\mathbf v$, that’s the best you can do. If $\mathbf n$ was a unit vector, you can normalize $\mathbf v-\mathbf v'$, but that gives you $\pm\mathbf n$. To resolve the sign ambiguity you’ll need to know if $\mathbf v$ is on the same side of the hyperplane as the original normal.