More generally, there are the special linear conformal transformations SL(2,R) associated with the differential operators
$S_{-1}f(z)=\exp\left(a\frac d{dz}\right)f(z)=f(z+a)$
$S_{0}f(z)=\exp\left(bz\frac d{dz}\right)f(z)=f(e^b z)$
$S_{1}f(z)=\exp\left(cz^{2}\frac d{dz}\right)f(z)=f\left(\frac z{1-cz}\right)$
The $z^{m+1}\frac d{dz}$ (m=-1,0,1) are a representation of a subgroup of the infinite Witt Lie algebra associated with the Virasoro algebra, and their exponential maps can be used to construct Möbius, or linear fractional, transformations.
For more info (combinatorics, generalizations), see my notes "Mathemagical Forests" (pages 13-15) at my little "arxiv".
Also refer to this question at Physics Forum.
(Update) Another way to look at the the scaling operator is
$S_{0}f(z)= exp[(e^t-1):zd/dz:]f(z)=exp[t\phi_{.}(:zd/dz:)]f(z)=exp(tzd/dz)f(z)$
where $(:zd/dz:)^n=z^n(d/dz)^n$ and $(\phi_{.}(x))^n=\phi_{n}(x)$ is the n’th Bell/Touchard/exponential polynomial with the exponential generating function $exp[(e^t-1)x]=exp[t\phi_{.}(x)]$.
Edit 2/2014: Also more simply, $exp[(a-1):zd/dz:]f(z)=f(a·z)$. And, the next logical extension is to treat $a$ as an umbral variable, i.e., $a^n=a_n$, as Blissard did.
Edit 6/2014: An equivalent op, when acting on fcts. analytic at the origin, is $exp(a:xD_{x=0}:)$.
Applying the last two ops. with $a$ an umbral variable to $exp(x)$ gives the Euler or binomial transformation for exp. generating fcts., which can then be related to the Euler transform for ordinary generating functions through the Borel-Laplace transform. Then evaluating at $x=1$ gives the Euler summation for a series. To me, the differential ops. make these relations transparent.