I'm not even sure what field of math this would be, and Googling "symmetry" and "functions" doesn't reveal what I'm looking for.
Basically I want to find all $\{\hat{A}\}$ other than the identity operator for a particular $\psi$ such that
$$\hat{A}\psi = \psi$$
Is this generally possible?
(I mean, if I wanted to get more specific I could say $\hat{A}$ is a bounded linear operator and $\psi$ is normalized and so on... but I just need a starting point for this kind of problem.)
EDIT: I'll try to give a little more information on why/what I'm trying to do here. Basically, I'm trying to extract symmetries from a function (like a wavefunction if you're familiar with physics) in order to reduce the information necessary to represent it.
For example, consider a spherically symmetric function $f$. One way to represent it is in the Cartesian basis as $$f(x,y,z) = e^{-\left(x^2 + y^2 + z^2\right)}$$
However, if you recognize that some kind of rotation operator applied to this function leaves it unchanged, you can rewrite it in the basis of spherical coordinates as:
$$f(r) = e^{-r^2}$$
Thus a function of 3 dimensions has been reduced to a function of 1 dimension by extracting a symmetry from the system. I'm trying to identify all such operators, so I can change the basis of the function into one that gives it a minimal representation.
There are going to be a lot of these operators. For instance, if $\psi = (1,0)^T$, then $\hat A$ of the form
$$ \hat A = \begin{pmatrix} 1 & x \\ 0 & y \end{pmatrix} $$
for any $x,y$ will work.
In more generality, if you're asking this in a vector space, let $\psi, e_2, e_3, \dots$ be a basis, then as long as $\hat A \psi = \psi$, you are completely free to choose $v_j = \hat A e_j$ however you want. There are a lot of such choices.