What classes of functions are closed under all rescalings? Let us denote by the symbol $\mathcal{G}$, a group of functions $f: \mathbb{R} \rightarrow \mathbb{R}$ (with the composition operation) that is additionally closed under all affine change of variables of the form (homothety):
$$ h(x) = mx, m>0$$
In other words, I would like the following property to hold for any affine maps $h$ (of the above form):
$$hGh^{-1} \in G$$
Intuitively such a group is a group of functions that is invariant under all rescalings.
A simple example of such a group is the group of fractional linear transformations (FLT) (with real coefficients), namely the group $\mathcal{S}$ consisting of that $f(x) = \frac{ax+b}{cx+d}$ where $a,b,c,d \in R$.
My questions are:


*

*Do such groups have a name?

*What is the classification of all such groups? (with the properties of $f'(x) >0$ and $f\in C^3$ if possible)

*Is there a general way of constructing such groups or putting this question in a general context?
Thank you in advance to all those who respond,
E(up)lio M.
.
 A: Let $G$ be any group of functions.  Let $H$ be the group of all compositions of functions of the form $f(mx)/m$ where $m>0$ and $f\in G$.  Then $H$ is a group of the sort you're looking for, and all such groups arise in this way.
A: If $G$ contains the linear group consisting of all functions of the form $h_m(x) = mx$ then obviously it satisfies your conditions.
Conversely, by the definition of the derivative,
(*) $lim_{m \to \infty} (h_m^{-1} f h_m(x)) = f'(0) x$
So the closure $\bar G$ of your group (in some appropriate topology) contains all linear functions of the form $h_m$ where $m = f'(0)$ for $f \in G$. So if you add to your conditions that the group $G$ be closed, and that the set of derivative values $f'(0)$, $f \in G$ consists of all positive real numbers, then $G$ does indeed contain the linear group.
A: For what its worth, there seem to be $2^{2^{2^{\aleph_0}}}$ many such groups (or at
least it is consistent with ZFC that there are this many). It is
easy to see the somewhat lesser claim that there must be at least
$2^{2^{\aleph_0}}$ many such groups, since we may simply close a
single bijective function under composition and conjugation by
affine functions. Since there are only continuum many affine
functions, this leads to a group of size at most continuum, and so
at most continuum many functions can lead to the same group this
way. But there are $2^{\frak{c}}=2^{2^{\aleph_0}}$ many
bijections. For the larger claim of $2^{2^{2^{\aleph_0}}}$,
consider the forcing to add $2^{\frak{c}}$ many mutually generic
bijections $f:\mathbb{R}\to\mathbb{R}$, without adding reals. The
groups generated by any subset of these functions in the
corresponding forcing extension, closing under composition and
conjugation by affine functions, will be different by mutual
genericity, and so there will be
$2^{2^{\frak{c}}}=2^{2^{2^{\aleph_0}}}$ many different such groups
in the forcing extension. I expect that one can get rid of this
forcing argument by more a more careful counting, and just prove
outright that there are $2^{2^{2^{\aleph_0}}}$ many such groups.
For example, all one needs is a family of $2^{\frak{c}}$ many
bijective functions $\mathbb{R}\to\mathbb{R}$, such that omitting
any one of them $f$ and closing the remaining functions under
composition and conjugation by affine functions, does not generate
$f$; that is, a maximal independent family for the generating
process in your question. In this case, every subset of that
family will generate distinct groups with your property, giving
$2^{2^{2^{\aleph_0}}}$ many such groups. The forcing argument
shows that it is consistent with ZFC to have such a large
independent family, but probably one can just prove it outright.
