$\newcommand\R{\mathbb R}\newcommand\la\lambda$It is well known and easy to see that the rotationally invariant product of two probability measures on $\R$ has to be a Gaussian (or Dirac) measure; see e.g. this answer.

This appears to make the conjecture below somewhat plausible.

Let $\mu$ be any probability measure on $\R^2$ with a finite nonzero covariance matrix. Let $\mu_1:=\mu$. For each natural $n$, consider the following three-step procedure:

1. let $$\la_n:=\mu_n^{(1)}\otimes\mu_n^{(2)},$$ where $\mu_n^{(1)}$ and $\mu_n^{(2)}$ are the marginals of $\mu_n$;
2. let $$\nu_n:=\int_0^{2\pi}\la_n R_t\,dt,$$ where $\la_n R_t$ is the pushforward measure obtained from $\la_n$ by the rotation about the origin through angle $-t$;
3. let $\mu_{n+1}$ be obtained by rescaling the probability measure $\nu_n$ so that the covariance matrix of $\mu_{n+1}$ be the unit matrix.

Conjecture: $\mu_n$ converges weakly (as $n\to\infty$) to the standard Gaussian measure on $\R^2$.

Is this conjecture true?

Comment: Perhaps Step 3, the rescaling, is not essential. Of course, if we have the convergence to a (nondegenerate) Gaussian measure without Step 3, then we have such a convergence with Step 3 as well.