Does there exist a probability distribution such that product of two IID variables is a uniform distribution? What about 3 IID variables? Product Distribution gives us a way to find the distribution of product of two IID random variables. This is in some sense the inverse question. If we know that $Z = X \times Y$ is uniformly random variable in the interval $[-a,a]$ then do there exist random variables X and Y (both identically distributed and independent) such that $Z = X \times Y$? Also do there exist $3$ IID random variables $W, X, Y$ such that $Z = X \times Y \times W$ ?

This question boils down to possible solution(s) of the following integral equation:
$$\int_{-\infty}^{+\infty}f(x)f\left(\frac{z}{x}\right)\frac{1}{|x|}dx=\frac{1}{2a}\left(u(z+a)-u(z-a)\right)$$
where $u(z)$ is the unit-step function and $a$ is a positive constant.

Maybe the moment-generating function can help you:
Suppose $X_1, \dots, X_n$ are iid and $Z := X_1 X_2 \cdots X_n$ is uniformly distributed on $[-a,a]$.
If I have not made a calculation error, the moments $m_k := E[Z^k], k \in \mathbb{N},$ should be given by
$$
m_k = \begin{cases} \frac{a^k}{k+1} &\text{ if $k$ is odd} \\ 0 &\text{ if $k$ is even} \end{cases}
$$
Since the $X_1, \dots, X_n$ are iid, we obtain
$$
m_k = E[Z^k] = E[X_1^k]E[X_2^k]\cdots E[X_n^k]=E[X_1^k]^n.
$$
If you can show that the function
$$
M(t) := 1 + \sum_{k=1}^\infty \frac{t^k \sqrt[n]{m_k}}{k!} = 1 + \sum_{k=1}^\infty \frac{t^{2k} \sqrt[n]{\frac{a^{2k}}{2k+1}}}{(2k)!}
$$
is well-defined, then this is the moment-generating function of the random variable $X_1$ that you are looking for.
Whether this is a 'known' distribution I do not know. But the moments should also give you some qualitative information about $X_1$.
