Figuring symmetries of a differential operator from its eigenfunctions

[JPaquim]

So, I understand that the derivative operator, D=\frac{d}{dx} has translational invariance, that is: x \rightarrow x - x_0, and its eigenfunctions are e^{\lambda t}. Analogously, the theta operator \theta=x\frac{d}{dx} is invariant under scalings, that is x \rightarrow \alpha x, and its eigenfunctions are x^\lambda. Taking logarithms and exponentials, I have constructed a sequence of operators and their respective eigenfunctions, all with the property that \{L(\frac{d}{dx})\}f^\lambda(x)=\lambda f^\lambda(x). I've taken a picture and attached it to this post.

My guess is that associated with every single one of these operators is some symmetry, some sort of coordinate transformation x \rightarrow f(x) under which the operator is invariant. For the x\log x \frac{d}{dx} operator, its invariant under x \rightarrow x^k, by inspection. How can I figure out what sort of symmetry a given operator has, given its eigenfunctions?

Physically, symmetries are associated with conservation laws. For a system whose differential equations are governed by this sort of differential operators, what sort of conserved quantities should I expect?

 

[The_Duck]

$\frac{d}{dx}$ is associated with translational symmetry. An infinitesimal translation is produced by acting on a function with the operator $1 + \epsilon \frac{d}{dx}$, with $\epsilon$ infinitesimal.

$x \frac{d}{dx}$ is associated with scale invariance. An infinitesimal rescaling is produced by acting on a function with the operator $1 + \epsilon x \frac{d}{dx}$, with $\epsilon$ infinitesimal.

Presumably your other operators $O$ can be associated with symmetry transformations with an infinitesimal transformation being implemented by $1 + \epsilon O$?

 

[JPaquim]

Ok, I agree with you. How can I figure out the "finite" version of the transformation from its infinitesimal counterpart?