Divergent Energies and Analytical Continuation - Two questions on the inverted harmonic oscillator and the inverted double well

I have two questions on the general topic of energy potentials that diverge at infinity.

First of all, the inverted harmonic oscillator. I found this post on Physics SE, Inverted Harmonic oscillator.
The answer from fellow user Mazvolej states that

"<…> The QHO does not permit analytic continuation, because it’s energies and wavefunctions depend not on $omega$, but on |$omega$|. Thus, their dependence on $omega$ is not analytic and $omega$ cannot be simply replaced by $iomega$.<…>".

I totally fail to see how the energies of the QHO depend on |$omega$|. Will they not depend on $omega^2$, which is analytic?

The paper Mavzolej links, Inverted Oscillator indeed shows that the naïve analytical continuation from ω to iω does not work, but I do not understand fully why.

Second question, I am trying to understand the reasoning in Anharmonic Oscillator. II A Study of Perturbation Theory in the Large Order

The authors consider a double well potential

$$frac{x^2}{4}+lambdafrac{x^4}{4}$$

When $lambda > 0$ bound states exist and the QHO energy spectrum is just perturbed by the $x^4$ term dominating at infinity.
When $lambda < 0$ the energy at infinity diverges to $-infty$.
They approach the problem by considering the function $E^k(lambda)$, where $k$ stand for the index of the energy eigenvalue, and considering its analytical continuation, as $lambda$ is rotated from the positive to the negative real axis.

The first obstacle for me are the boundary conditions at infinity they set to solve Schroedinger’s equation.
I cite their reasoning on page 1623, which is totally delphic to me (they denote $-lambda = epsilon$):

*"<…> At $x = +infty$ the boundary conditions are somewhat complicated <…>. It would appear that any linear combination of outgoing and incoming waves $exp(pm epsilon^{1/2} x^3/6)$ would suffice. However, we recall that the analytical continuation of the energy levels into the complex plane is accomplished by simultaneously rotating $x$ into te complex $x$ plane. When $arg lambda = pi$, the sector in which the boundary condition $lim _{lvert x rvert to infty} psi(x) = 0$ applies is given by $-frac{1}{3} pi < arg(pm x)<0$. Thus it is necessary to pick that asymptotic behaviour which vanishes exponentially if the argument of $x$ lies between $0 ^circ$ and $-60^circ$. Hence, $Psi(x)$ must obey the boundary condition $$ Psi(x) sim frac{const}{x} exp(-i epsilon^{1/2} x^3/6)$$ as $x to +infty$ <…>".

I am not so sure I grasp this.
Here are my thoughts.

For $x to +infty$, the quartic term will dominate and I understand the asymptotic behaviour $$exp(pm iepsilon^{1/2} x^3/6)$$ is expected, when $lambda$ is real and negative.
The analytical continuation could be achieved by keeping $lambda$ real and negative, and rotating $x$ (by why do they say, "simultaneously"? Is $lambda$ also rotated? Why both?).

By writing $x = lvert x rvert exp(i theta)$ and substituing in the asymptotic expression (with the $-$ sign, the one the authors claim to be the right boundary condition
$$exp(- epsilon^{1/2} frac{x^3}{6})$$
I get
$$exp(- iepsilon^{1/2} frac{lvert x rvert^3}{6} (cos 3theta + i sin 3theta)) sim exp(- epsilon^{1/2} frac{lvert x rvert^3}{6} sin 3theta) $$
where in the last step the oscillatory component was discarded.
The solution will hence decay to $0$ if $ 0 > theta > -pi/3$, and also this seem to match what they say.

Now, the wave function has to vanish for positive $lambda$ and $x to +infty$, as in the QHO.
This should be equivalent to keeping $lambda$ negative and rotating $x$. But if $x$ is rotated by $pi$, the asymptotic behaviour will not be exponentially decaying to $0$.

I would be grateful for any hint on my mistake.