Is $f''(x)+x^{-2}f(x)=0 $ translation invariant? Let $f:\mathbb R \to \mathbb C$ satisfies the following differential equation (DE):
$$f''(x)+x^{-2}f(x)=0  $$
Questions:

If $f$ satisfies the above (DE),  then can we say $f_{x_0}(x)=f(x-x_0)$ ($0\neq x_0 \in \mathbb R$) also satisfies the above (DE)?
If $f$ satisfies the above (DE), then what can we say about $f$? (Is there any, method to obtain $f$ from the (DE)?)

 A: No, the DE is not translation invariant in the sense you specified. 
Fix $a \in \mathbb R$, $a \ne 0$, and let $g(x) = f(x-a)$.
Then $g(x) = f(x-a) \implies g'(x) = f'(x-a) \implies g''(x) = f''(x-a)$.
Then 
$$f''(x)+x^{-2}f(x)=0, \forall x \in \mathbb R$$
$$\implies f''(x-a)+(x-a)^{-2}f(x-a)=0, \forall x \in \mathbb R$$
$$\implies g''(x)+(x-a)^{-2}g(x)=0, \forall x \in \mathbb R$$
which is not the same as
$$g''(x)+x^{-2}g(x)=0, \forall x \in \mathbb R$$
A: $$\frac{d^2f}{dx^2}+\frac{f}{x^2}=0$$
Of course, the ODE isn't translation invariant,  but the ODE is expansion or contraction invariant. The change of variable $\quad x=at\quad$ leads to :
$$\frac{d^2f}{dt^2}+\frac{f}{t^2}=0$$
This mean that the ODE is invariant (Of course, the solution isn't invariant : the coefficients in the solution change with $a$, but the analytic form of the solution doesn't change).
Analytic solving of the ODE :
Change of variable $\quad x=e^X \quad\to\quad \frac{d^2f}{dX^2}-\frac{df}{dX}+f=0$
$$f=c_1e^{X/2}\cos\left(\frac{\sqrt{3}}{2}X\right)+c_2e^{X/2}\sin\left(\frac{\sqrt{3}}{2}X\right)$$
$$f(x)=c_1x^{1/2}\cos\left(\frac{\sqrt{3}}{2}\ln|x|\right)+c_2x^{1/2}\sin\left(\frac{\sqrt{3}}{2}\ln|x|\right)$$
The change of $x$ to $ax$ changes $\quad \ln|x|\quad$ to $\quad\ln|x|+\ln|a|.\quad$ A few trigonometric transformations leads to the same form of solution, but with different constants $C_1$ , $C_2$ instead of $c_1$ , $c_2$.
