The normalizing factors are the same.

Let us take two matrices $R \in \mathbb{R}^{m\times n}, S \in \mathbb{R}^{l \times n}$ such that $R^TR = S^TS$. Now, $\bar{R}$ can be obtained by multiplying $R$ with an appropriate diagonal matrix $D_R$ from the right: $\bar{R} = RD_R$, where $$D_R^2 = \sum_{k=1}^n I_n^{(k)} R^TRI_n^{(k)},$$ where $I_n^{(k)}$ is an $n\times n$ matrix with zeros everywhere, except for the element at $(k,k)$, where it has a 1. Since $R^TR = S^TS$, we have $D^2_R = D^2_S$, thus (being diagonal matrices with non-negative elements) $D_R = D_S$.

The normalizing factors are the same.

Let us take two matrices $R \in \mathbb{R}^{m\times n}, S \in \mathbb{R}^{l \times n}$ such that $R^TR = S^TS$. Now, $\bar{R}$ can be obtained by multiplying $R$ with an appropriate diagonal matrix $D_R$ from the right: $\bar{R} = RD_R$, where $$D_R^2 = \sum_{k=1}^n I_n^{(k)} R^TRI_n^{(k)},$$ where $I_n^{(k)}$ is an $n\times n$ matrix with zeros everywhere, except for the element at $(k,k)$, where it has a 1. Since $R^TR = S^TS$, we have $D^2_R = D^2_S$, thus (being diagonal matrices with non-negative elements) $D_R = D_S$.

Hence, $$\bar{R}^T\bar{R} = (RD_R)^T(RD_R) = D_RR^TRD_R = D_SS^TSD_S = \bar{S}^T\bar{S}.$$

The normalizing factors are the same.

Let us take two matrices $R \in \mathbb{R}^{m\times n}, S \in \mathbb{R}^{l \times n}$ such that $R^TR = S^TS$. Now, $\bar{R}$ can be obtained by multiplying $R$ with an appropriate diagonal matrix $D_R$ from the right: $\bar{R} = RD_R$, where $$D_R^2 = \sum_{k=1}^n I_n^{(k)} R^TRI_n^{(k)},$$ where $I_n^{(k)}$ is an $n\times n$ matrix with zeros everywhere, except for the element at $(k,k)$, where it has a 1. Since $R^TR = S^TS$, we have $D^2_R = D^2_S$, thus (being diagonal matrices with non-negative elements) $D_R = D_S$.

I don't think $\bar{R}^T\bar{R} = \bar{S}^T\bar{S}$ would generally be true. In your example you used specially shaped $l \times 1$ matrices. For such matrices, $R$ and $D_R$ commute and $\bar{R}^T\bar{R} = \bar{S}^T\bar{S}$ follows.

The normalizing factors are the same.

Let us take two matrices $R \in \mathbb{R}^{m\times n}, S \in \mathbb{R}^{l \times n}$ such that $R^TR = S^TS$. Now, $\bar{R}$ can be obtained by multiplying $R$ with an appropriate diagonal matrix $D_R$ from the right: $\bar{R} = RD_R$, where $$D_R^2 = \sum_{k=1}^n I_n^{(k)} R^TRI_n^{(k)},$$ where $I_n^{(k)}$ is an $n\times n$ matrix with zeros everywhere, except for the element at $(k,k)$, where it has a 1. Since $R^TR = S^TS$, we have $D^2_R = D^2_S$, thus (being diagonal matrices with non-negative elements) $D_R = D_S$.