So a hypothetical question in which the hypothetical breaks a law of physics can always be problematic, since you need to decide how to break the law of physics, and that choice cannot be justified by physics---nevertheless, ppl will attack your choice based on it violating physics, which it has to do.

tl;dr: any comments about ignoring a law of physics will ignored with equal vigor.

Your supposition is correct, for an atom with a nuclear mass $M$, one can posit:

$$ \frac 1 2 \le \frac{m_e} M \le 294 $$

which starts at the exotic atom positronium, and tops out at Oganesson-294.

That means the nucleus is not stationary.

Labeling the two particles' positions $\vec r_N$, $\vec r_e$, we have a hamiltonian (in the Coulomb approximation...and likewise for the position):

$$ H(\vec r_N, \vec r_e) = \frac{p^2_N}{2M} + \frac{p^2_e}{2m_2} + V(|\vec r_N - \vec r_e|) $$

which is not fun.

We note that the position dependence can be expressed in term of the vector:

$$ \vec r \equiv \vec r_e - \vec r_N $$

moreover, it only depends on the magnitude:

$$ r= |\vec r| $$

(which means we have spherical symmetry).

Step 1 is to defined the center of mass affine position:

$$ \vec R = \frac{M\vec r_N + m_e \vec r_e} {M+m_e} $$

with conjugate momentum $\vec P$.

Step 2 is define the reduced mass, $\mu$, associated with the coordinate vector $\vec r$:

$$ \frac 1 {\mu} = \frac 1 M + \frac 1 {m_e} $$

The Hamiltonian is then:

$$ H = \frac{P^2}{2(M+m_e)} + \frac{p^2}{2\mu} + V(r) $$

which contains a free-particle term:

$$H_{CoM} \equiv \frac{P^2}{2(M+m_e)} $$

and a pure Coulomb term:

$$H \equiv \frac{p^2}{2\mu} + V(r) $$

where:

$$ V(r) = \frac 1 {4\pi\epsilon_0} \frac{Ze^2} r $$

At this point, we ignore the center-of-mass motion, and solve for the orbit (classical) or hydrogen-like atom (quantum).

As you point out in the post: yes, the nucleus undergoes 'reciprocal' motion. This means it classical is also in an elliptical orbit about the barycenter (aka: CoM), and in quantum mechanics, is also in an orbital with a wave function $\Psi_{nlm}(\vec r_N)$ that has the same functional form as the atomic orbitals $\psi^{(e)}(\vec r_e)$.

Nevertheless, supper imposing the two motions is messy, so we call the atomic orbital

$$ \psi_{nlm}(\vec r) $$

and compute it with mass $\mu$.

The original question: "how are the two ellipses related" is that they are related by dilation symmetry, where the scale factor is a ratio of the two masses.

In the case of positronium:

$$\frac{m_e}{M} = \frac{m_e}{m_e}=1 $$

and they are identical, which for the largest atom:

$$ \frac{m_e}{M} \approx \frac{m_{AMU}/1823}{294m_{AMU}} \approx \frac{1}{536,000}$$

so the size of the ellipse (or orbital) of the nucleus is:

$$ a_N \approx 0.1\,{\rm fm} = \frac{R_p}{12} $$

where:

$$ R_p = 0.88\,{\rm fm} $$

is the proton radius.

Regarding the hypothetic part: a classical orbit radiates EM waves because charges are accelerated. Since they electron moves at:

$$ v \approx \alpha c $$

in a radius:

$$ a_0 =\lambda_C/\alpha $$

where $\alpha \approx 1/137$ is the fine structure constant, and

$$\lambda_C = \frac h {m_ec} \approx 4\times 10^{-13} \,{\rm m} $$

is the Compton wavelength of the electron, we can ball park the acceleration:

$$ a = \frac{v^2} r = \frac{\alpha^2 c^2}{\lambda_C/\alpha}$$

$$ a = \frac{m_e}{h}\big(\alpha c\big)^3 \approx 10^{22}\, g $$

Wow. That is big, and that is for $Z=1$. I'll leave it as an exercise to estimate the result for $Z=118$.

Quantum mechanics of course precludes this state from radiating.

So a hypothetical question in which the hypothetical breaks a law of physics can always be problematic, since you need to decide how to break the law of physics, and that choice cannot be justified by physics---nevertheless, ppl will attack your choice based on it violating physics, which it has to do.

tl;dr: any comments about ignoring a law of physics will ignored with equal vigor.

Your supposition is correct, for an atom with a nuclear mass $M$, one can posit:

$$ \frac 1 2 \le \frac{m_e} M \le 294 $$

which starts at the exotic atom positronium, and tops out at Oganesson-294.

That means the nucleus is not stationary.

Labeling the two particles' positions $\vec r_N$, $\vec r_e$, we have a hamiltonian (in the Coulomb approximation...and likewise for the position):

$$ H(\vec r_N, \vec r_e) = \frac{p^2_N}{2M} + \frac{p^2_e}{2m_2} + V(|\vec r_N - \vec r_e|) $$

which is not fun.

We note that the position dependence can be expressed in term of the vector:

$$ \vec r \equiv \vec r_e - \vec r_N $$

moreover, it only depends on the magnitude:

$$ r= |\vec r| $$

(which means we have spherical symmetry).

Step 1 is to defined the center of mass affine position:

$$ \vec R = \frac{M\vec r_N + m_e \vec r_e} {M+m_e} $$

with conjugate momentum $\vec P$.

Step 2 is define the reduced mass, $\mu$, associated with the coordinate vector $\vec r$:

$$ \frac 1 {\mu} = \frac 1 M + \frac 1 {m_e} $$

The Hamiltonian is then:

$$ H = \frac{P^2}{2(M+m_e)} + \frac{p^2}{2\mu} + V(r) $$

which contains a free-particle term:

$$H_{CoM} \equiv \frac{P^2}{2(M+m_e)} $$

and a pure Coulomb term:

$$H \equiv \frac{p^2}{2\mu} + V(r) $$

where:

$$ V(r) = \frac 1 {4\pi\epsilon_0} \frac{Ze^2} r $$

At this point, we ignore the center-of-mass motion, and solve for the orbit (classical) or hydrogen-like atom (quantum).

As you point out in the post: yes, the nucleus undergoes 'reciprocal' motion. This means it classically is also in an elliptical orbit about the barycenter (aka: CoM), and in quantum mechanics, it is also in an orbital with a wave function $\Psi_{nlm}(\vec r_N)$ that has the same functional form as the atomic orbitals $\psi^{(e)}(\vec r_e)$.

Nevertheless, super-imposing the two motions is messy, so we call the atomic orbital

$$ \psi_{nlm}(\vec r) $$

and compute it with mass $\mu$.

The original question: "how are the two ellipses related" is that they are related by dilation symmetry, where the scale factor is a ratio of the two masses.

In the case of positronium:

$$\frac{m_e}{M} = \frac{m_e}{m_e}=1 $$

and they are identical, while for the largest atom:

$$ \frac{m_e}{M} \approx \frac{m_{AMU}/1823}{294m_{AMU}} \approx \frac{1}{536,000}$$

so the size of the ellipse (or orbital) of the nucleus is:

$$ a_N \approx 0.1\,{\rm fm} = \frac{R_p}{12} $$

where:

$$ R_p = 0.88\,{\rm fm} $$

is the proton radius.

Regarding the hypothetic part: a classical orbit radiates EM waves because charges are accelerated. Since they electron moves at:

$$ v \approx \alpha c $$

in a radius:

$$ a_0 =\lambda_C/\alpha $$

where $\alpha \approx 1/137$ is the fine structure constant, and

$$\lambda_C = \frac h {m_ec} \approx 4\times 10^{-13} \,{\rm m} $$

is the Compton wavelength of the electron, we can ball park the acceleration:

$$ a = \frac{v^2} r = \frac{\alpha^2 c^2}{\lambda_C/\alpha}$$

$$ a = \frac{m_e}{h}\big(\alpha c\big)^3 \approx 10^{22}\, g $$

Wow. That is big, and that is for $Z=1$. I'll leave it as an exercise to estimate the result for $Z=118$.

Quantum mechanics of course precludes this state from radiating.

(As an aside: as $Z\alpha \rightarrow 1$ this estimate runs into relativity problems as $v\rightarrow c$, and the naive radius gets down to the Compton wavelength)