Let ${\bf x} \in \mathbb{R}^n$ and ${\bf z} \in \mathbb{R}^m$ denote two real-valued and bounded random variables.
Assume that the following expectation values are the same for all moments of $\bf x$ (for all possible ${\bf k}$): $E\left[\prod_{i=1}^n x_i^{k_i}\right] = E\left[f_{\bf k}({\bf x} ,{\bf z})\right]$, where ${\bf k}=(k_1,...,k_n)$ and $k_i\in\mathbb{N}_0$. Here, $f_{\bf k}$ denote real valued scalar functions. In general, $f_{\bf k}({\bf x},{\bf z})\neq \prod_{i=1}^n x_i^{k_i}$.
Then, can we conclude $E\left.\left[\prod_{i=1}^n x_i^{k_i}\right|{\bf z}\right] = E\left.\left[f_{\bf k}({\bf x},{\bf z})\right|{\bf z}\right]$?
I understand that the conditional expectations have to exist because all $E\left[\prod_{i=1}^n x_i^{k_i}\right]$ exist because $\bf x$ is bounded. Also, all mixed moments exist and uniquely defined the probability distribution of $\bf x$. But can the equality in the assumption transfer to the conditional expectation values?
No we cannot. The following counter example disproves it.
Let $X$ and $Z$ be i.i.d. $\mathbb{R}^n$ valued bounded random vectors. Define $f_{\bf k}(X,Z) = \prod_{i=1}^n Z^{k_i}$, then by the assumption $X\sim Z$ we get $$\mathbb{E}[\prod_{i=1}^n X_i^{k_i}] = \mathbb{E}[\prod_{i=1}^n Z^{k_i}]=\mathbb{E}[f_{\bf k}(X,Z)]$$ although usually $\prod_{i=1}^n X_i^{k_i}\neq \prod_{i=1}^n Z^{k_i}$. By independence we have that $$ \mathbb{E}[\prod_{i=1}^n X_i^{k_i} \: | \: Z] = \mathbb{E}[\prod_{i=1}^n X_i^{k_i}]$$ and by measurability we have that $$\mathbb{E}[\prod_{i=1}^n Z_i^{k_i} \: | \: Z] = \prod_{i=1}^n Z_i^{k_i}$$ and equality between the two conditional expectations would imply that $\prod_{i=1}^n Z_i^{k_i}=\mathbb{E}[\prod_{i=1}^n Z_i^{k_i}]$ almost surely, which is usually not the case except for trivial/degenerate cases.