Are conformal maps between Riemannian manifolds real-analytic? This is a cross-post.

Let $M,N$ be oriented smooth ($C^{\infty}$) $n$-dimensional Riemannian manifolds, and let $f:M \to N$ be a smooth orientation-preserving weakly* conformal map.
Do there exist real-analytic structures on $M,N$ that make $f$ real-analytic?

I only assume the metrics are $C^{\infty}$.
Every smooth manifold has a unique real-analytic structure (up to diffeomorphism) compatible with its smooth structure.
A  reasonable starting point would be to know whether every $C^{\infty}$ conformal map between real-analytic manifolds with real-analytic metrics is real-analytic. (I don't know a reference for that; anyway, what I am asking seems harder).
*A weakly conformal map is a map whose differential at every point is either conformal or zero. (This is equivalent to $df^Tdf =(\det df)^{\frac{2}{n}} \, \text{Id}_{TM}$).
Motivation:
I am trying to understand if smooth weakly conformal maps whose differential vanishes at a point are constant (for dimensions $n \ge 3$). This seems to be the case for analytic maps, hence my interest in the possible analyticity of such maps.

For the Euclidean case, this follows directly by Liouville's theorem:
For $n=2$, every such map is complex-analytic. Let $\Omega \subseteq \mathbb{R}^n$ be an open subset, $n \ge 3$, and let $f:\Omega \to \mathbb{R}^n$ be a smooth conformal map. By Liouville's theorem, $f$ is of the form
$$ f(x)=b+\alpha\frac{1}{|x-a|^\epsilon}A(x-a),$$
where $A$ is an orthogonal matrix, and $\epsilon \in \{0,2\}, b \in \mathbb{R}^n,\alpha \in \mathbb{R},a \in \mathbb{R}^n \setminus \Omega$.
So, up to translations and dilations,
$ f(x)=\frac{A x}{|x|^2}$ (where $ 0 \notin \Omega$) which is real-analytic as a multiplication of two analytic maps. ($1/x^2$ is analytic on $\mathbb{R} \setminus \{0\}$).
 A: As essentially follows from your argument in the comment to Piotr Hajlasz's answer (now deleted), for $f\colon M\to N$ to be a counterexample, it cannot be a diffeomorphism. Because then you could just pick any analytic structure on $N$, pull it back to $M$ by $f$ and make $f$ trivially analytic.
With that in mind, one can think of counterexamples already in 1-dimension. For instance let $(N,g_N) = (M,g_M) = (\mathbb{R},dx^2)$, where $dx^2$ is the metric that appears flat with respect to some global coordinate $x$. Let
$$
  f(x) = \begin{cases}
    (x+\epsilon) e^{-1/(x+\epsilon)^2} & x<-\epsilon \\
    0 & -\epsilon\le x \le \epsilon \\
    (x-\epsilon) e^{-1/(x-\epsilon)^2} & \epsilon < x
  \end{cases} .
$$
It is locally a diffeomorphism if you consider it as a map from $\mathbb{R}\setminus [-\epsilon,\epsilon]$ to $\mathbb{R}\setminus \{0\}$. But there is no pair of analytic structures on $\mathbb{R}$ that would make $f\colon \mathbb{R} \to \mathbb{R}$ analytic. If $f$ were analytic (with respect to any analytic atlas), then it could not vanish on any open set, while not being globally zero. But since we are in 1-dimension, any map is weakly conformal. So then $f\colon M \to N$ is weakly conformal, but can never be made analytic.
Higher dimensional examples could be generated from this one by using rotational symmetry and basically reducing to the 1-dimensional case.
A: This is not an answer. Just a possible vector of attack for this problem.
According to the Theorem 11.4.6 of Harmonic Morphisms Between Riemannian Manifolds by Paul Baird and John C. Wood, every non-constant weakly conformal map between manifolds of the same dimension $n\geq 3$ is a conformal local diffeomorphism. The automorphism group of $(M, [g])$ is a Lie group whose Lie algebra is the algebra of conformal Killing vector fields. 
According to the Theorem 3.8 of Normal BGG solutions and polynomials every conformal Killing vector field is given by a polynomial expression (of degree at most 2) in normal coordinates and normal frame. Given a Cartan geometry $(\pi: \mathcal{G} \to M, \omega \in \Omega^1(\mathcal{G}, \mathfrak{g})$, these normal coordinates and frames are given by flow flow of $\omega^{-1}(X)$ and by projection of that flow through $\pi.$
