Is $g(x)= \frac{x_1^2}{r_1^2}+\frac{x_2^2}{r_2^2}- c$ a unique solution of $E[g(Z)\mid M=\mu]=0, \forall \mu \in \text{ellipse } C$ for $Z$ Gaussian Let $Z \in \mathbb{R}^2$ be an i.i.d. Gaussian vector with mean $M$ where $P_{Z\mid M}$ is its distribution.
Let $g: \mathbb{R}^2 \to \mathbb{R}$ and consider the following equation:
$$
E[g(Z)\mid M=\mu]=0,  \forall \mu \in C,
$$
where $C=\{\mu: \frac{\mu_1^2}{r_1^2}+\frac{\mu_2^2}{r_2^2}=1 \}$ for some given $r_1,r_2 > 0$. That is, $C$ is an ellipse.
It is not difficult to verify (see this question, see also Edit 3) that a solution to this equation is given by
$$
g(x)= \frac{x_1^2}{r_1^2}+\frac{x_2^2}{r_2^2}- c,
$$
where $c=\frac{1}{r_1^2}+\frac{1}{r_2^2}+1$.
In fact any function $g_a(x)= a g(x)$ for any $a \in \mathbb{R}$ is a solution.
Question: Is $g$ a unique solution up to a multiplicative constant?
Edit: If we need to make an assumption on  the class of allowed functions $g$. Let us assume that $g$'s are bounded by a quadratic monomial (i.e., for every $g$ there exists $a$ and $b$ such that $g(x) \le a \|x \|^2 +b$).
Edit 2: If you want to avoid expectation notation. Everything can be alternatively written as
$$
\iint g(z)\frac{1}{2 \pi} e^{-\frac{\|z-m\|^2}{2}} \, {\rm d} z=0, \, m\in C.
$$
From here, one can see that this question is about a convolution.
Edit 3: To see that $g(x)$ is a solution we use that the second moment of Gaussian is given by  $E[Z_i^2\mid M_i=\mu_i]=1+\mu_i^2$, which leads to
\begin{align}
E\left[\frac{Z_1^2}{r_1^2}+\frac{Z_2^2}{r_2^2}- c\mid M=\mu \right]&=  \frac{E[Z_1^2\mid M_1=\mu_1]}{r_1^2}+\frac{  E[Z_2^2\mid M_2=\mu_2]}{r_2^2}-c\\[6pt]
&=\frac{1+\mu_1^2}{r_1^2}+\frac{  1+\mu_1^2}{r_2^2}-c\\[6pt]
&=\frac{1}{r_1^2}+\frac 1 {r_2^2}+1-c,
\end{align}
where in the last step we used that $\mu$ is on the ellipse.
Edit 4: The comment below shows that the solution is not unique when an ellipse is a circle.  However, I would still like to know the answer for a general ellipse.
 A: This will not answer the question as stated, but whoever asks such a question may be interested in what I say here.
First, a random variable with a "Gaussian" or "normal" distribution need not have variance $1,$ so I would have stated what the variance is.
In standard terminology, the question is asking which unbiased estimators of zero this parameterized family of probability distributions has (where the ellipse is the parameter space).
A family of probability distributions that admits no unbiased estimators of zero is called "complete", so the question is how far this family is from being complete.
The family of all univariate Gaussian distributions, whose parameter space may be taken to be the half-plane $\{(\mu,\sigma) : \mu\in\mathbb R,\, \sigma>0\},$ is complete, and that fact is essentially the same thing as the one-to-one nature of the two-sided Laplace transform.
A corollary of the completeness of this family of all Gaussian distributions is that with an i.i.d. sample, the sample mean has a smaller variance than every other unbiased estimator of the population mean. For example, think about how to prove that the sample median has a smaller variance than the sample mean. With this result, you get that without computing the variance of the sample median.
