NOTE: I have first asked this on physics.stackexchange, they advised me to ask on math.stackexchange
The operator $X$ is called the position operator in physics with it's conjugate being the momentum operator: $$[X,P]=i$$ ($\hbar=1$). Eigenstates of the position operator is known as quadrature/position states: $$X|x\rangle=x |x\rangle$$. But what about the eigenstates of the $X^N$ operator? Here $N$ is an integer greater than one. For example for $N=2$ we can say that superposition of the position operators are one of the eigenstates:
$$ X^2 (|x\rangle+|-x\rangle)=x^2 (|x\rangle+|-x\rangle)$$
As you can check. But also the position eigenstate is a $X^2$ eigenstate: $X^2|x\rangle =x^2 |x\rangle$.
Is there a general method or formula to find the eigenstates of $X^N$ in terms of the eigenstates of $X$?
If the eigenstates of $X$ form a basis (which is true for position operator, but doesn't have to be true in the general case), then you can look for the eigenstate in this basis.
Let $|f\rangle = \int f(x)|x\rangle dx$ and let us assume that $$ X^N\int f(x)|x\rangle dx = \lambda \int f(x)|x\rangle dx$$ from the linearity of $X^N$ it means that $$ \int f(x)x^N|x\rangle dx = \lambda \int f(x)|x\rangle dx $$ $$ \int f(x)(x^N-\lambda)|x\rangle dx = 0 $$ which can only be satisfied if $f(x) = 0$ for $x^N \neq \lambda$. Since $x\in\mathbb R$, that means that for $N$ odd we have $f(x)=0$ for $x \neq \sqrt[N]\lambda$, and for $N$ even we have $f(x)=0$ for $x \notin \{\sqrt[N]\lambda, -\sqrt[N]\lambda\}$.
In conclusion, the eigenstates of operator $X^{2N+1}$ are all of the form $\alpha|\sqrt[2N+1]\lambda\rangle$, and eigenstates of $X^{2N}$ are all of the form $\alpha|\sqrt[2N]\lambda\rangle+\beta|-\sqrt[2N]\lambda\rangle$