A monic degree-5 polynomial in $\mathbb{Z}_{5}[x]$ for which all elements of $\mathbb{Z}_{5}$ are roots I found is $(x^5 - x)$ since
$f(1) \equiv$ 0(mod 5) $f(2) \equiv$ 0(mod 5) $f(3) \equiv$ 0(mod 5) $f(4) \equiv$ 0(mod 5) $f(5) \equiv$ 0(mod 5).
I am wondering, however, if there could be more than one answer. I noticed that $(x^5 - x)$ is a difference and can be factored but I don't know if that is a good conjecture.
Hint: if their were two then their difference is a nonzero polynomial of degree $< 5$ with $5$ roots, contra: a polynomial in a field (or domain) has no more roots than its degree.
Remark $ $ More generally, by above if $f(x) = 0\,$ for all $x$ then $\deg f \ge 5.\,$ Let $\, g = x^5-x.\,$ Note $h = f\bmod g = f - q\,g\,$ also has $\,h(x) = 0$ for all $x$ and $\,\deg h < 5\,$ so $\, h = 0,\, $ so $\,f = q\,g.\,$ Thus any other polynomial having the same set of roots is a multiple of the minimal degree polynomial $g$.
Remark $ $ This is a special case of the fact that ideals in Euclidean domains are principal - generated by any element of minimal Euclidean value(= degree here).