Jacobian of $A (A^\top X A)^{-1} A^\top$ Let $A\in\mathbb{R}^{n\times m}$, $n\geq m$, be a full column rank matrix, and consider the function
\begin{align}
f&\colon \mathbb{R}^{n\times n} \to \mathbb{R}^{n\times n}\\
& X\mapsto A (A^\top X A)^{-1} A^\top,
\end{align}
where $\bullet^\top$ denotes transposition.
Assuming that $(A^\top X A)^{-1}$ exists, I'm interested in the computation of the Jacobian matrix of $f$, i.e.
$$\tag{1}\label{a}
\mathbf{J}[f] = \left[\frac{\partial f(X)}{\partial X_{ij}}\right]\in\mathbb{R}^{n^2\times n^2}.
$$
I know that there exists a closed form expressions for the Jacobian of the inverse, namely $\mathbf{J}[X^{-1}]=-(X^{-\top} \otimes X^{-1})$ (see e.g. here, page 5). Hence, I wonder whether a similar closed-form expression can be derived for \eqref{a}. 
Thanks in advance.
 A: Given $\mathrm A \in \mathbb R^{n \times m}$, matrix-valued function $\mathrm F : \mathbb R^{n \times n} \to \mathbb R^{n \times n}$ is defined as follows
$$\mathrm F (\mathrm X) := \mathrm A \left( \mathrm A^{\top} \mathrm X \mathrm A \right)^{-1} \mathrm A^{\top}$$
Hence,
$$\mathrm F (\mathrm X + h \mathrm V) = \mathrm A \left( \mathrm A^{\top} (\mathrm X + h \mathrm V) \mathrm A \right)^{-1} \mathrm A^{\top} = \cdots = \mathrm F (\mathrm X) -  h \mathrm A \left( \mathrm A^{\top} \mathrm X \mathrm A \right)^{-1} \mathrm A^{\top} \mathrm V \mathrm A \left( \mathrm A^{\top} \mathrm X \mathrm A \right)^{-1} \mathrm A^{\top}$$
Thus, the directional derivative of $\mathrm F$ in the direction of $\mathrm V$ at $\mathrm X$ is the matrix-valued function
$$- \mathrm A \left( \mathrm A^{\top} \mathrm X \mathrm A \right)^{-1} \mathrm A^{\top} \mathrm V \mathrm A \left( \mathrm A^{\top} \mathrm X \mathrm A \right)^{-1} \mathrm A^{\top}$$
Making $\mathrm V = \mathrm e_i \mathrm e_j^{\top}$, we obtain
$$\partial_{x_{ij}} \mathrm F (\mathrm X) = - \mathrm A \left( \mathrm A^{\top} \mathrm X \mathrm A \right)^{-1} \mathrm A^{\top} \mathrm e_i \mathrm e_j^{\top} \mathrm A \left( \mathrm A^{\top} \mathrm X \mathrm A \right)^{-1} \mathrm A^{\top} = \color{blue}{- \mathrm F (\mathrm X) \, \mathrm e_i \mathrm e_j^{\top} \mathrm F (\mathrm X)}$$
which is a multiple of the outer product of the $i$-th column and $j$-th row of $\mathrm F (\mathrm X)$.
Vectorizing the directional derivative, we obtain
$$\mbox{vec} \left( - \mathrm A \left( \mathrm A^{\top} \mathrm X \mathrm A \right)^{-1} \mathrm A^{\top} \mathrm V \mathrm A \left( \mathrm A^{\top} \mathrm X \mathrm A \right)^{-1} \mathrm A^{\top} \right) = \color{blue}{- \left( \mathrm A \left( \mathrm A^{\top} \mathrm X^{\top} \mathrm A \right)^{-1} \mathrm A^{\top} \otimes \mathrm A \left( \mathrm A^{\top} \mathrm X \mathrm A \right)^{-1} \mathrm A^{\top} \right)} \mbox{vec} (\mathrm V)$$
A: For typing convenience, define the matrix
$$\eqalign{
\def\B{B^{-1}}
\def\p{\partial}
\def\qq{\qquad\qquad}
\def\vc{\operatorname{vec}}
\def\grad#1#2{\frac{\p #1}{\p #2}}
B &= A^TXA \\
dB &= A^T\;dX\;A \\
d\B &= -\B\;dB\;\B \;=\; -\B A^T\;dX\;A\B \\
}$$
and use it to write the function and calculate its differential
$$\eqalign{
F &= A\B A^T \\
dF &= A\;d\B\,A^T \;=\; -F\;dX\;F \qq \\
}$$
from which one may obtain the componentwise matrix-valued gradients
$$\eqalign{
\grad F{X_{ij}} &= -F\,E_{ij}\,F \qq\qq\qq \\
}$$
or a vectorized matrix-valued gradient
$$\eqalign{
\grad {\vc(F)}{\vc(X)} &= -{F^T\otimes F} \qq\qq\quad \\
}$$
or a tensor-valued gradient
$$\eqalign{
\grad FX &= -F\,{\cal E}\,F^T \qq\qq\qq \\
}$$
where
${\cal E}$ is a fourth-order tensor whose components can be written in terms of Kronecker delta symbols as
$\,{\cal E}_{ijk\ell} = \delta_{ik}\,\delta_{j\ell}\;$
while
$\;E_{ij}$ is a matrix whose components are all zero except for
the $(i,j)$ component which is equal to one.
