= Solution
Full column rank makes $B=A^TA$ positive definite: for nonzero $v$, $v^TBv=\|Av\|^2>0$. Let $B=R^TR$ be its unique upper-triangular <Cholesky decomposition> with positive diagonal. Defining $Q=AR^{-1}$ gives
$$
Q^TQ=R^{-T}A^TAR^{-1}=I,\qquad A=QR.
$$
Thus the skinny <QR decomposition> exists. If $A=\widetilde Q\widetilde R$ is another such decomposition, orthonormality gives $A^TA=\widetilde R^T\widetilde R$. Cholesky uniqueness forces $\widetilde R=R$, and then $\widetilde Q=AR^{-1}=Q$. Therefore \b[the skinny factorization with positive diagonal is unique]. This is the <thin QR factorization from Cholesky decomposition>; no inverse of the rectangular $Q$ is assumed.
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