When can an arbitrary function be put into this form? A problem I've been trying to address but not been able to get very far with is to devise a method to check whether or not some function $q(x)$ can be written like
$$
q(x) = \varphi^{-1}[1+\varphi(x)].
$$
Or, put another way, "given a function $q(x)$, how do we decide whether or not there exists a solution $\varphi(x)$ to $\varphi\circ q - \varphi = 1$?".
Yet another way to recast the problem is, "for a given $q(x)$, how do we know whether or not one can find a function $f$ so that $e^{f(x)\frac{d}{dx}}x = q(x)$?".
At the very least, $q$ must be invertible, which is apparent from it's definition ($q^{-1}(x) = \varphi^{-1}[\varphi(x) - 1]$). As a matter of practicality, one can make definitions for $\varphi$ without any loss of generality to arrive at other equivalent forms for $q$, for example $\varphi\rightarrow\ln\sigma\Rightarrow q(x)\rightarrow\sigma^{-1}[e\sigma(x)]$.
Actually finding $\varphi$ for a given $q$ is something I've been able to do a few times via infinite matrices (with much sweat and tears), and any information on that process would be fun for me to hear about too. I'm sure there are already far more general existence and uniqueness theorems that answer my question...I can recall coming across some that consider similar things, at the very least, but I still have trouble absorbing most mathematical literature at this point in my education. Any pointers in the right direction, i.e. things I could be considering, any meaningful discussion is greatly appreciated. 
 A: Actually, I had a thought. While this is definitely not the whole story, it certainly seems true that if 
$$
q: I \longrightarrow\, I
$$
for some $I\in\mathbb{R}$, and we specify some point $\phi(x_0) = \phi_0,\,x_0\in I$, then we have some type of solution on $I$. This was easy to see after I tried a super brute-force method of a table of values. If by dotting the $x$-axis with each pair of points $(x,q(x))$ on a plot of $\phi(x)$, one fills up the entire interval $I$:
$$\forall x'\in I,\,\,q(x')\in I$$
Then we have a corresponding value for $\phi$ at each point $x'$ in $I$. There is lots of room for trouble, though, i.e. if $q$ has a fixed point. If, though for each $x'\in I$ there is a unique $q(x')$, so that $q$ is one-to-one on $I$, then it is possible to construct $\phi$ by starting with $\phi(q(x_0)) = 1 + \phi(x_0) \equiv 1 + \phi_0$, and then plugging in $\phi(q(q(x_0)))$, etc. for each pair to get a resulting pair $(\phi(q(x_i)),\phi(x_i) + 1)$ to plot on the $\phi$-axis. That is,
$$
(q)^n(x_0) = \phi^{-1}(n+\phi_0)
$$
Where $(q)^n$ is $n$-fold composition of $q$, i.e. $(q)^3(y) = q(q(q(y)))$. 
We can sort of say, then
$$
\phi^{-1}(x) \sim (q)^{x-\phi_0}(x_0)
$$
This actually holds for a few $q$, and gives correct answers when verifying with $q(x) = \phi^{-1}(1+\phi(x))$ or equivalently $e^{\partial_\phi}x = q(x)$. There is the issue of the set $\{(q)^n(x_0), \forall n \in \mathbb{Z}\}$ covering the interval $I$ in an optimal, 'disjoint' fashion, which worries me when it comes to proofs, but as far as actually doing what I want to do, the formula $(q)^n(x_0) = \phi^{-1}(n+\phi_0)$ works just fine when one can analytically expand $n$ to real values, i.e. $(q)^n(x_0) \equiv f(x_0,n)$.
So, then, I humbly submit a (certainly incomplete, if not incorrect altogether) first guess of some conditions on $q$: 
The equation
$$
(\phi\circ q)(x) - \phi(x) = 1
$$
Has a solution, $\phi(x)$, for all $x \in I$, if $q$ is one-to-one on $I$, and has no fixed points on $I$ (perhaps no attractive fixed points near $I$?).
