Real valued function concatenation as group operator? For which sets? Can we define function concatenation of normal functions $$x\in \mathbb R\\x\to f(x)\in\mathbb R$$ as group operator ?
The identity element would be $f(x) = x$, I suppose
Function inverse can be defined as $f^{-1}(x)$ s.t. $f^{-1}(f(x)) = 0$.
In group setting I suppose our group's inverse element could simply be this function inverse?
These are the two seemingly uncontested properties which fit well.

But... what about associativity? $(f \circ g) \circ h = f \circ (g \circ h)$. Are we sure this will be satisfied?
And what about closure? What families of functions would guarantee this? One example should be polynomials, but I am quite sure that inverse element would often not exist within polynomials...
 A: You only need closure under composition and inverses as composition of functions is always associative. 

You may prove this as follows: Let $f:X\to Y$, $g:Y\to Z$, $h:Z\to W$ be functions with $X,Y,Z,W$ non-empty sets. Then, for any $x\in X$, we have
$$((h\circ g)\circ f)(x)=(h\circ g)(f(x))=h(g(f(x)))=h((g\circ f)(x))=(h\circ(g\circ f))(x)$$
and so $(h\circ g)\circ f$ and $h\circ(g\circ f)$ are pointwise equal and thus equal as functions.
Note that this "proof" is really just analysis the nature of composition as we just shift the brackets around; there is no hidden meaning; composition is inherently associative.

An example would the space of all continuous functions from $\mathbb R$ to $\mathbb R$ which are invertible, or in other words bijective. Composition of continuous functions gives you a continuous function and the composition of two bijective functions is bijective.

Another classical example is the group of automorphism of an algebraic structure, e.g. for a vector space $V$ the set of all bijective linear maps $\phi: V\to V$.
