First sample $X\sim\text{Poisson}(n)$. Then toss a $n$-faced fair die $X$ times independently, letting $Y_k$ denote the total number of times face number $k\in\{1,\ldots,n\}$ has showed up. It is clear that $$X=\sum_{k=1}^nY_k$$
whatever the random realization of $X$ may be.
Further, $Y_1,\ldots,Y_n$ are independent $\text{Poisson}(1)$ r.v. Indeed, for any $a_1,\ldots,a_n\in\Bbb Z_+$ we have, with $N:=a_1+\cdots+a_n$,
\begin{align*}
\Bbb P\Bigl(Y_1=a_1,\ldots,Y_n=a_n\Bigr)&=\Bbb P\Bigl(Y_1=a_1,\ldots,Y_n=a_n\:\Big|\: X=N\Bigr)\cdot\Bbb P(X=N)\\[.4em]
&=\binom{N}{a_1,\ldots,a_n}\left(\frac1n\right)^{\!N}\cdot{\mathrm e}^{-n}\frac{n^N}{N!}\\[.4em]
&=\left(\frac{\mathrm e^{-1}}{a_1!}\right)\cdots\left(\frac{\mathrm e^{-1}}{a_n!}\right)\\[.4em]
&=\Bbb P\Bigl(\text{Poisson}(1)=a_1\Bigr)\cdots\Bbb P\Bigl(\text{Poisson}(1)=a_n\Bigr).
\end{align*}
Formal construction of $(Y_1,\ldots,Y_n)$. As stated in my earliest comment, we need a bit of extra independent randomness for this construction. So, suppose there exists a Uniform$(0,1)$ random variable $U$ independent of $X$. (If it does not exist, say we augment the probability space by taking its product with $((0,1),\mathcal B[0,1),\text{Leb})$.)
Then $(Y_1,\ldots,Y_n):=f(X,U)$ where $f\colon\Bbb Z_+\times(0,1)\to\mathbb(\Bbb Z_+)^n$ is the measurable map such that for any $y_1,\ldots,y_n\in\mathbb Z_+$, we have $(y_1,\ldots,y_n)=f(x,u)$ if, for all $1\le k\le n$, $y_k$ is the number of occurrences of digit $k$ in the first $x$ digits of $u$ when written in base $n$.
Best Answer
There is no such distribution except the trivial one where $X_i=0$. If $\phi$ is the characteristic function then we get $\phi^{n}\equiv \phi$ for all $n$. If $|\phi (t)| <1$ this gives (by letting $n \to \infty$) $\phi (t)=0$. But $\phi (0)=1$ and $\phi$ is continuous so we must have $|\phi (t)| =1$ for $|t|$ sufficiently small. It is easy to see from this $X_1=0$ almost surely.