Consider a probability distribution on $d$ events, with the probabilities $p_j$ gathered in a vector $\vec p\in \mathbb{R}^d$. For natural numbers $k$, define the $k$-th moment of surprisal as $$m_k(\vec p) = \sum_j p_j (-\log(p_j))^k.$$ (When one of the $p_j=0$, we define $0(-\log(0))^k=0$ by continuity.) The moments only depend on $\vec p$ up to permutations and knowing the moments of surprisal for all $k$ determines the distribution uniquely up to permutations.
I currently believe that $m_1,\ldots,m_{d-1}$ are in fact sufficient to determine the vector $\vec p$ up to permutations. (In fact, I currently believe that knowing any subset of $d-1$ different moments is sufficient, since they ought to be sufficiently independent to determine the distribution up to permutations ($d-1$ parameters).) However, I am not able to prove this.
Note that for a random variable on $d$ events taking values $x_j$, where all the $x_j$ are distinct, knowing the moments $k=0,\ldots,d-1$ is in fact sufficient to reconstruct the probability distribution. However, we cannot imply take $x_j=-\log(p_j)$, since the reconstruction of the $p_j$ in the case of a random variable requires knowing the $x_j$ in advance, while here they are determined by the $p_j$.
It is not too difficult to see that for distributions where all the $p_j$ are distinct and strictly positive, that the function $\vec p \mapsto m_k(\vec p)$ is locally invertible. However, it's not clear to me how to establish global invertability (up to permutations) even in this case.
Any insight in regard to this, including literature references (or a counter-example!) would be very much appreciated!