Let $m\times n$ matrix be $A$. Rank of $A$ is $n$, where $n \leq m$, so $A$ is a full-column matrix.
The rank of $A$ is $n$ if and only if there exists an invertible $m\times m$ matrix $X$ and $n \times n$ invertible matrix $Y$ that satisfy
$XAY=\begin{pmatrix} I_n\\ 0_{(m-n)\times n} \end{pmatrix}$
More specifically, the above fact is demonstrated in https://en.wikipedia.org/wiki/Rank_(linear_algebra)
I'm wondering if there exists $X$ and $Y$ such that $X'X=I_m$ and $Y'Y=I_n$, for any $A$. If it doesn't hold for any $A$, then what condition of $A$ will lead $X$ and $Y$ to satisfy $X'X=I_m$ and $Y'Y=I_n$?
This is readily addressed using singular value decomposition (SVD). $\underline {\overline {\bf{X}} }$ and $\underline {\overline {\bf{Y}} }$ will be orthogonal if and only if the $n$ singular values of $\underline {\overline {\bf{A}} }$ are all unity, which then implies that $${\underline {\overline {\bf{A}} } ^T}\underline {\overline {\bf{A}} } = {\underline {\overline {\bf{I}} } _{\left[ n \right]}}$$
The additional details: $\underline {\overline {\bf{A}} }$ can always be written as $$\underline {\overline {\bf{A}} } = {\underline {\overline {\bf{X}} } ^T}\underline {\overline {\bf{S}} } \,{\underline {\overline {\bf{Y}} } ^T}$$ where $\underline {\overline {\bf{X}} }$ and $\underline {\overline {\bf{Y}} }$ are the desired orthogonal matrices. Hence $$\underline {\overline {\bf{X}} } \,\underline {\overline {\bf{A}} } \,\underline {\overline {\bf{Y}} } = \left( {\underline {\overline {\bf{X}} } \,{{\underline {\overline {\bf{X}} } }^T}} \right)\underline {\overline {\bf{S}} } \,\left( {{{\underline {\overline {\bf{Y}} } }^T}\underline {\overline {\bf{Y}} } } \right) = \underline {\overline {\bf{S}} }$$ Since the singular values of the SVD are unique (other than their ordering), for your situation you require $$\underline {\overline {\bf{S}} } = \left[ {\begin{array}{*{20}{c}} {\left[ {\begin{array}{*{20}{c}} {{s_1}}& \cdots &0\\ \vdots & \ddots & \vdots \\ 0& \cdots &{{s_n}} \end{array}} \right]}\\ {\underline {\overline {\bf{0}} } } \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{{\underline {\overline {\bf{I}} } }_{\left[ n \right]}}}\\ {\underline {\overline {\bf{0}} } } \end{array}} \right]$$ Then $${\underline {\overline {\bf{A}} } ^T}\underline {\overline {\bf{A}} } = \underline {\overline {\bf{Y}} } \,{\underline {\overline {\bf{S}} } ^T}\left( {\,\underline {\overline {\bf{X}} } \,{{\underline {\overline {\bf{X}} } }^T}} \right)\underline {\overline {\bf{S}} } \,{\underline {\overline {\bf{Y}} } ^T} = \underline {\overline {\bf{Y}} } \left[ {\begin{array}{*{20}{c}} {{{\underline {\overline {\bf{I}} } }_{\left[ n \right]}}}&{\underline {\overline {\bf{0}} } } \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\underline {\overline {\bf{I}} } }_{\left[ n \right]}}}\\ {\underline {\overline {\bf{0}} } } \end{array}} \right]{\underline {\overline {\bf{Y}} } ^T} = \underline {\overline {\bf{Y}} } \,{\underline {\overline {\bf{Y}} } ^T} = {\underline {\overline {\bf{I}} } _{\left[ n \right]}}$$