Holomorphic maps on $\mathbb{R}^{n}$ (for n not necessarily even)

Question

Edit according to the comment of user36931 I remove the "motivation" from the previous version and I add an statement to the first question

We consider the following two classes of smooth maps on $\mathbb{R}^{n}.$ ( $n$ is not necessarily even):

$HP_{1}$: $\;$A smooth map $f:\mathbb{R}^{n} \to \mathbb{R^{n}}$ satisfies property $HP_{1}$ if the pull back metric $f^{*}(g)$ is a conformal metric for every conformal metric $g$ on $\mathbb{R}^{n}.$ ( By conformality I mean $g=e^{h} \sum dx_{i}^{2})$

$HP_{2}$: $\;$A smooth map $X$ on $\mathbb{R}^{n}$ with coordinates $X=(P_{1},P_{2},\ldots,P_{n})$ satisfies $HP_{2}$ if the space of harmonic functions is invariant under the derivational operator $U \mapsto X.U=\sum P_{i} \partial U/ \partial x_{i}$

Questions:

1. Assume $n=2k$, identify $\mathbb{R}^{2k}$ with $\mathbb{C}^{k}$. Let $f$ satisfies $HP_{1}$. Is $f$ a holomorphic map on $\mathbb{C}^{k}$? The same question for $HP_{2}$?

2. Are the above two properties, equivalents?(A map is $HP_{1}$ iff it is $HP_{2}$?)

Answer 1

This is an extended comment. The first property $HP_1$ simply says that $f$ is conformal. It is not true that holomorphic maps of $C^n=R^{2n}$ are conformal, except when $n=1$. In fact there are very few conformal maps in $R^n$ for $n\geq 3$: they are only Mobius transfomrations (compositions of reflections).