It is well-known that no two-dimensional point lattice contains a regular pentagon. (See for example http://mathworld.wolfram.com/LatticePolygon.html.) The same is true for lattices in $\mathbb{R}^n$, simply because any such polygon would lie in a two-dimensional sublattice. If we relax the requirement that the pentagon lie in a plane, we can easily find a closed path of 5 equal length sides: for example, in $\mathbb{R}^5$ with the integer lattice, we can use the path $$(1,0,0,0,0)\to(0,1,0,0,0)\to(0,0,1,0,0)\to(0,0,0,1,0)\to(0,0,0,0,1).$$ (These are the vertices of the standard 4-simplex.) The same construction, of course, works to find a nonplanar equilateral lattice "$n$-gon" for any $n$, as the vertices of the standard $n+1$-simplex in the integer lattice. We can improve this to a lattice in $\mathbb{R}^{n-1}$ by restricting to the plane $x_1+\cdots+x_n=1$. My question is if this is the best we can do, dimension-wise. So is it possible, for example, to find a nonplanar equilateral lattice pentagon in some lattice in $\mathbb{R}^3$? What can be said about the minimal dimension of lattice containing a nonplanar equilateral lattice $n$-gon?

(Note: This was originally posted on math stackexchange here, but nobody answered it.)