Calculate $\frac{\partial}{\partial \mathbf{A}}\lVert \mathbf{A}^{\top}\mathbf{AX}-\mathbf{X} \rVert _{F}^{2}$ with $\mathbf{A}\in\mathbb{R}^{M\times N}$ and $\mathbf{X}\in\mathbb{R}^{N\times D}$, and find $\mathbf{A}$ satisfying $$\frac{\partial}{\partial \mathbf{A}}\lVert \mathbf{A}^{\top}\mathbf{AX}-\mathbf{X} \rVert _{F}^{2}=\mathbf{0}_{M\times N}$$
My Efforts
I know that $\frac{\partial}{\partial \mathbf{B}}\lVert \mathbf{B} \rVert _{F}^{2}=2\mathbf{B}$, and am not sure that if $\frac{\partial}{\partial \mathbf{A}}(\mathbf{A}^\top\mathbf{A})=2\mathbf{A}$ (I get these from The Matrix Cookbook). How to get the final derivative, and solve $\frac{\partial}{\partial \mathbf{A}}\lVert \mathbf{A}^{\top}\mathbf{AX}-\mathbf{X} \rVert _{F}^{2}=\mathbf{0}_{M\times N}$?
Stripping away the norm, you're attempting a least-squares solution of $$A^TAX = X \quad\implies\quad A^TA = XX^+$$ QR decomposition yields $$\eqalign{ QR &= XX^+ \\ (QR)^TQR &= (XX^+)^T(XX^+) \\ R^TQ^TQR &= (XX^+)(XX^+) \\ R^TR &= (XX^+) \\ }$$ Therefore $A=R$ is one possible solution. To obtain the desired dimensions, drop the last $(M-N)$ rows of $R$.
Since there was no requirement for $A$ to be triangular, other (non-triangular) solutions are possible using different factorizations, e.g. SVD or EVD decompositions.