Consider a map
$$ \varphi:A\to \exp\bigg(\frac{A}{\log(B)}\bigg) $$
where $A,B\in {PSL(2,\Bbb Z)},$ (modular group).
Solve $\varphi(B)=B.$
The solution is $A=\log^2(B).$ However make a change of variables and define $A:=\log^2(\mathfrak C)$ for $\mathfrak C$ an infinite set of matrices in the modular group. So $\mathfrak C=\{C_k\}$ for $C_k\in PSL(2,\Bbb Z)$ for $\forall k=1,2,\cdot\cdot\cdot$. This is compact notation. One could also write out the infinite number of equations $-$ one equation for each $C\in\mathfrak C.$ I will write a few out just to be clear:
$$ \varphi_1(B)=\exp\bigg( \frac{\log^2(C_1)}{\log(B)} \bigg)=B$$
$$ \varphi_2(B)=\exp\bigg( \frac{\log^2(C_2)}{\log(B)} \bigg)=B $$
$$... $$
So in compact notation one has:
$$\varphi(B)=\exp\bigg(\frac{\log^2(\frak C)}{\log(B)}\bigg)=B$$
Solve $\varphi(B)=B.$
I've been trying to classify the solution space for hours, but I need to take a break and I thought I'd ask for some hints.
More of an extended comment:
We end up with:
$$C_1=B$$
$$C_2=B$$
$$ \cdot\cdot\cdot $$
This is due to manipulating the equations algebraically.
It's interesting to note that there is something of substance here. We have a smoothly varying space of objects, in this case, matrices in the modular group. We could have a different group of objects. The main idea is:
If one varies $B$ via $\varphi(B)$ then we get something like a continuously varying space of objects.