If $L^T L = R$ is the available Cholesky decomposition, then inverting both sides of the equation you get,
$$L^{-1}(L^{T})^{-1} = R^{-1} $$
And since transposition and inverse are interchangeable:
$$L^{-1}(L^{-1})^{T} = R^{-1} $$
So if you define $P = (L^{-1})^T$ this is your desired answer. In other words,
$$P^{T}P=R^{-1}$$
Let's use a naming convention where matrices and vectors are denoted
by upper and lower case Latin letters, respectively. Further, the symbol $\odot$ will denote the Hadamard product and $\otimes$ the Kronecker product.
For ease of typing, use $\{S,A,P\}$ instead of
$\,\{\Sigma,{\large\Lambda},{\cal P}\}\,$
and $\,X=f(\Sigma)\,$.
Then rewrite the problem using these conventions.
$$\eqalign{
S &= XX^T,\quad
A = I\odot X,\quad
V=A^{-1} \\
P &= VSV + A - I \\
}$$
Each of these matrices (except for $X$) is symmetric, and $(A,V,I)$ are diagonal.
Apply the vec operation ($K$ denotes the Commutation matrix)
$$\eqalign{
y &= {\rm vec}(I) \\
x &= {\rm vec}(X) \quad\implies\quad{\rm vec}(X^T) &\doteq Kx \\
a &= y \odot x
\;=\; {\rm Diag}(y)\,x
&\doteq Yx \\
da &= Y\,dx \\
\\
s &= (I\otimes X)Kx \;=\; (X\otimes I)\,x \\
ds &= \Big((I\otimes X)K+(X\otimes I)\Big)\,dx &\doteq N\,dx\\
\\
p &= (V\otimes V)s + a-y &\doteq Bs + a-y \\
&= (VS\otimes I)v + a-y &\doteq Hv + a-y \\
&= (I\otimes VS)v + a-y &\doteq Jv + a-y \\
}$$
Finally, calculate the differentials of $v$ and $p$
$$\eqalign{
dv &= {\rm vec}(-V\,dA\,V) = -(V\otimes V)\,da \\
&= -B\,da \;=\; -BY\,dx \\
\\
dp &= da + B\,ds + H\,dv + J\,dv \\
&= Y\,dx + BN\,dx - (H+J)BY\,dx \\
&= \Big(Y + BN - (H+J)BY\Big)\,dx \\
}$$
and the gradient with respect to $x$
$$\eqalign{
\frac{\partial p}{\partial x}
&= Y + BN - (H+J)BY \\
&= Y
+ (V\otimes V)\Big((I\otimes X)K+(X\otimes I)\Big)
- (VS\otimes I + I\otimes VS)(V\otimes V)Y \\
&= Y + (V\otimes VX)K+(VX\otimes V)
- \big(VSV\otimes V + V\otimes VSV\big)Y \\
}$$
Best Answer
$\Sigma^{-1} = T^{-1} (T^{*})^{-1} = R^*R\implies I= TR^*RT^*=\big(RT^*\big)^*\big(RT^*\big)\implies Q^* = RT^*$
for some unitary $Q$
Multiply each side on the left by $Q$ and on the right by $(T^*)^{-1}$ to get
$(T^*)^{-1} = QR$
That is, they are related by the unique QR factorization of $(T^*)^{-1}$ such that $R$ has positive diagonals (uniqueness follows since the Cholesky factorization is unique for PD matrices).