Closest conformal matrix to a given matrix $\newcommand{\SOn}{\operatorname{SO}_n}$
$\newcommand{\COn}{\operatorname{CO}_n}$
$\newcommand{\Sym}{\operatorname{Sym}_n}$
$\newcommand{\Skew}{\operatorname{Skew}_n}$
$\newcommand{\dist}{\operatorname{dist}}$
$\newcommand{\Sig}{\Sigma}$
$\newcommand{\sig}{\sigma}$
$\newcommand{\al}{\alpha}$
$\newcommand{\id}{\operatorname{Id}}$
Let $A \in \text{GL}_+$ be an $n \times n$ real matrix with positive determinant. Is there always a closest conformal matrix to $A$? (I measure the distance between matrices using the Euclidean distance). Is the closest matrix unique?
By using SVD we can restrict to the case where $A$ is positive diagonal . 
Edit:
Let $\Sig=\text{diag}(\sigma_1,\dots,\sigma_n)$ be diagonal with positive entries. Suppose all the $\sig_i$ are different, and that $\sig_1 < \sig_2 < \dots \sig_n$.
I prove below that if a minimizer exists, then it is $\bar \Sig=\frac{\sum_{i=1}^n\sigma_i}{n}\id.$
However, the question of existence of a minimizer still remains. Since the conformal group is not compact, it's non-trivial. We can restrict to given ball of course, but if it contains $0$ then we have a problem, since the intersection with the conformal group won't be closed.

Conjecture: A minimizer always exist and unique. Equivalently, for any positive diagonal matrix $\Sig$ its closest matrix is given by $\bar \Sig$.

I proved it for the case where the entries of $\Sig$ are all different, assuming the existence of a minimizer. (I can adapt the argument to the case of multiplicities but it's a bit cumbersome).

Proof the closest matrix is diagonal: (If someone finds an easier proof that would be nice).
We denote the (special) conformal group by $\COn$, i.e $$ \COn = \{Q \in M_n \, | \, Q^TQ=(\det Q)^{\frac{2}{n}}\id,\det Q >0  \}.$$
Let $\Sig=\text{diag}(\sigma_1,\dots,\sigma_n)$ be diagonal with positive entries. If a closest matrix $Q \in \COn$ exists then I prove below it must be diagonal (hence a scalar times the identity, since it's conformal). 
So, the problem is reduced to minimizing $\sum (\sigma_i - \sigma_{\rm new})^2$. Thus, the closest matrix is $$Q=Q(\Sig)=\bar \Sig=\frac{\sum_{i=1}^n\sigma_i}{n}.$$
In particular the closest matrix is unique.
Suppose $Q \in \COn$ satisfies $$ d( \Sig,Q) = \dist( \Sig  ,\COn), \tag{1}$$
and let $\al(t)=\Sig+t\left( Q-\Sig \right)$ be the minimizing geodesic from $\Sig$ to $Q$. Since a minimzing geodesic from a point to a submanifold intersects that submanifold perpendicularly, we get 
$$
\dot \al (1) = (Q- \Sig) \perp T_Q \COn
$$
The map $X \to Q^{-1}X$ is homothety of $\text{GL}_+$ (endowed with the Euclidean Riemannian metric), so it preserves orthogonality. Thus,
$$
\id-Q^{-1}\Sig=Q^{-1}(Q-\Sig) \in (T_{\id}\COn)^{\perp} \subseteq (\Skew)^{\perp}=\Sym,
$$
which implies $Q^{-1}\Sig \in \Sym$, or $$  Q^{-1}\Sig = \Sig (Q^{-1})^T. \tag{2}$$
Since $Q \in \COn$, $Q^TQ=(\det Q)^{\frac{2}{n}}\id$. So, $Q^{-T}=(\det Q)^{-\frac{2}{n}}Q$. Plugging this into $(2)$ we obtain
$$ Q^{-1}\Sig = \Sig (\det Q)^{-\frac{2}{n}}Q, $$ or equivalently
$$ \big( (\det Q)^{-\frac{1}{n}}Q \big)^{-1} \Sig = \Sig \cdot (\det Q)^{-\frac{1}{n}}Q \tag{3}.$$
Denote $P:=(\det Q)^{-\frac{1}{n}}Q$, and note that $P \in \SOn$. Then $(3)$ is equivalent to
$$ P^T\Sig=\Sig P. \tag{4}$$
It suffices to show $P$ is diagonal. 
$$
P^T\Sig = \begin{pmatrix}
  \sig_1P_{11} & \sig_2 P_{21}  & \cdots & \sig_nP_{n1} \\
\sig_1P_{12} & \sig_2 P_{22}  & \cdots & \sig_nP_{n2} \\
 \vdots  & \vdots& \vdots & \vdots & \\  
\sig_1P_{1n} & \sig_2 P_{2n}  & \cdots & \sig_nP_{nn} \\
\end{pmatrix} = \begin{pmatrix}
  \sig_1 P_{11} & \sig_1 P_{12}  & \cdots & \sig_1 P_{1n} \\
\sig_2P_{21} & \sig_2 P_{22}  & \cdots & \sig_2P_{2n} \\
 \vdots  & \vdots& \vdots & \vdots & \\  
\sig_nP_{n1} & \sig_n P_{n2}  & \cdots & \sig_nP_{nn} \\
\end{pmatrix} = \Sig P \tag{5}
$$
Looking at the first column of both matrices in $(5)$ and comparing (squared) norms we get 
$$
  \sig_1^2 \sum_{i=1}^n P_{1i}^2=\sig_1^2=\sig_1^2 \sum_{i=1}^n P_{i1}^2=\sum_{i=1}^n \sig_i^2 P_{i1}^2. \tag{6}
$$
Since we assumed $\sig_1$ is strictly smaller than all the other singular values, we deduce that $P_{i1} = 0$ for $i > 1$. Inserting this back to $(6)$, we get
$$
\sig_1^2 \sum_{i=1}^n P_{1i}^2=\sum_{i=1}^n \sig_i^2 P_{i1}^2=  \sig_1^2 P_{11}^2,
$$
So, we deduce $P_{1i}=0$ for $i >1$.
Now we continue in this way.
 A: $\newcommand{\SOn}{\operatorname{SO}_n}$
$\newcommand{\COn}{\operatorname{CO}_n}$
$\newcommand{\Sym}{\operatorname{Sym}_n}$
$\newcommand{\Skew}{\operatorname{Skew}_n}$
$\newcommand{\dist}{\operatorname{dist}}$
$\newcommand{\Sig}{\Sigma}$
$\newcommand{\sig}{\sigma}$
$\newcommand{\al}{\alpha}$
$\newcommand{\id}{\operatorname{Id}}$
Here is complete answer. There always exist a unique minimizer.
Let $\Sigma=\text{diag}(\sigma_1,\dots,\sigma_n)$ be diagonal and positive.  We want to solve
$$ \min_{Q \in \COn} \| \Sig-Q\|^2=\min_{Q \in \COn} \| \Sig\|^2+ \|Q\|^2-2\langle \Sig,Q \rangle.$$
Since $\Sig$ is constant, this is equivalent to solving
$$\min_{Q \in \COn}  F(Q), \, \text{ where } \, F(Q):=\|Q\|^2-2 \sum_{i=1}^n \sig_iQ_{ii}.$$
Any $Q \in COn$ can be written uniquely as $\lambda O$ for some $\lambda \in \mathbb{R}^+$, $O \in \SOn$. Since
$$ F(\lambda O)=\lambda^2n-2\lambda\sum_{i=1}^n \sig_iO_{ii},$$
it is obvious that
$$ F(\lambda O) \ge F(\lambda \id).$$ So, we are reduced to optimizing $\lambda$, i.e solve
$$ \min_{\lambda \in \mathbb{R}^+} F(\lambda \id)=\min_{\lambda \in \mathbb{R}^+} \lambda^2n-2\lambda\sum_{i=1}^n \sig_i.$$
By differentiating we get that $\lambda=\frac{\sum_{i=1}^n\sig_i}{n}$.
