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This is probably a standard fact but I cannot find it in my usual references. Let $G$ be one of the classic matrix Lie groups $\mathrm{U}(N), \mathrm{SU}(N), \mathrm{O}(N), \mathrm{SO}(N)$, equipped with the Haar measure $\mu$. One way to study the $t$-th moment of $\mu$ is by defining the matrix $$\Phi^{(t)}_\mu = \int_G U^{\otimes t} \otimes (U^*)^{\otimes t} \, \mu(dU),$$ where $U^*$ is the entrywise complex conjugate of $U$. At what value of $t$ do we have a complete characterization of the distribution? For example, in the scalar setting, a Gaussian distribution over $\mathbb{R}$ is completely characterized by its first two moments.

I believe an equivalent way to phrase the question is also: given some normalized measure $\nu$ on $G$, what is the smallest $t$ such that $\Phi^{(t)}_\mu = \Phi^{(t)}_\nu$ iff $\mu = \nu$?

Banach space fan
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The two formulations given in your question are in fact not equivalent: While the normal/Gaussian distribution is determined by the first two moments, you still have to check all moments if you want to find out whether the distribution is Gaussian! (and infinitely many are non-zero).

As far as I know, the Gaussian setting is not simpler then the generic setting where you want to find out whether two probability measures are equal by looking at (all of) their moments. Note that given that all moments agree, there are still pathological cases when the measures do not, see Hamburger moment problem.

I think both questions are interesting, but I do not know a resolution to the first one: Are all moments of the Haar measure determined by the moments up to a finite $t$?

Thus, let me answer the second question instead: Another way of looking at the "moment problem" is via characteristic functions, or in a more modern language, Fourier transforms of measures. For (finite Borel) measures $\nu$ on compact groups $G$, we can define a Fourier transform $\hat\nu$ by the collection of (finite-dimensional) operators, indexed by irreps $\tau$ of $G$: $$ \hat\nu(\tau) := \int_G \bar\tau(g) \,d\nu(g)\,. $$ Then, one can show that the mapping $\nu\mapsto\hat\nu$ is an isomorphism on the Banach space of finite Borel measures on $G$ (into a slightly weird space though), and hence $\nu=\mu$ iff $\hat\nu=\hat\mu$. This is yet another version of a fundamental result in harmonic analysis, the Peter-Weyl theorem.

Since the compact groups you mentioned have infinitely many irreps, we have to check $\hat\nu=\hat\mu$ on all of these. Moreover any irrep is contained in $U^{\otimes t} \otimes \bar{U}^{\otimes t}$ for sufficiently high $t$, implying that is indeed not enough to check equality of your moment operators for finite $t$.

A side remark: For finite $t$, you would get a unitary $t$-design, and these can in general even by finite sets (i.e. counting measure on those).

Markus Heinrich
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