By using appropriate vector norms it is possible to extract an equilateral triangle from a two-dimensional lattice, and a regular pentagon from a four-dimensional one. By "regular" I mean, in each case, equilateral and equiangular; as we will see in the four-dimensional case the figure (living in four-dimensional space) is not a planar regular pentagon.
The triangular case
The matrix
$M=\begin{pmatrix}0&-1\\1&-1\\\end {pmatrix}$
has the property that if we take successive powers its column vectors cycle among three values:
$M^1=\begin{pmatrix}0&-1\\1&-1\\\end {pmatrix}$
$M^2=\begin{pmatrix}-1&1\\-1&0\\\end {pmatrix}$
$M^3=\begin{pmatrix}1&0\\0&1\\\end {pmatrix}$
and of course the next powers repeat this pattern as $M^3$ is the identity. With that in mind we might suppose that the three points represented by these vectors
$(0,1),(1,0),(-1,-1)$
should somehow form an equilateral triangle.
Plotting on an ordinary Cartesian coordinate plane gives no evidence of any such equality. But suppose that instead of putting the positive $y$ axis at $90°$ from the positive $x$ axis, we were to make that angle $120$°. Using the Law of Cosines for triangles we can render distances from the origin in terms of a modified version of the Euclidean norm:
$||u||=||(u_1,u_2)||^2=u_1^2-u_1u_2+u_2^2 =\frac14(u_1+u_2)^2+\frac34(u_1-u_2)^2$
We see that in this norm all three points are equidistant from the origin, namely all the vectors have norm $1$.
But what about angular displacements? It is not difficult to see that for any vectors $u,v$ the above norm satisfies the relation
$||u+v||^2+||u-v||^2=2(|u|^2+|v|^2)$
and so we can define an associated inner product:
$(u\cdot v)=\frac14(||u+v||^2-||u-v||^2)=u_1v_1+u_2v_2-\frac12(u_1v_2+u_2v_1)$
When this is applied to the three coordinate vectors given above, we find that they all satisfy
$||u||=1,(u\cdot v)=-\frac12.$
In other words, with the right norm all three points are equidistant from the origin and at equal angles relative to each other, thus we have rendered an equilateral triangle using only two dimensions instead of three!
Left: The points $(1,0),(0,1),(-1,-1)$ plotted in ordinary Cartesian coordinates do not form an equilateral triangle. Right: Modifying the coordinates so that lengths match a suitable metric renders the triangle equilateral.
The fourth dimension
For the pentagonal case we must start with a matrix whose powers give a 5-cycke of vectors. Such a matrix must have a characteristic polynomial equal to the fifth cyclotomic polynomial, which is degree $4$, and so we move to $\phi(5)=4$ dimensions. Analogously to the two-dimensional case, we have:
$M=\begin{pmatrix}0&0&0&-1\\1&0&0&-1\\0&1&0&-1\\0&0&1&-1\\\end{pmatrix}$
$M^2=\begin{pmatrix}0&0&-1&1\\0&0&-1&0\\1&0&-1&0\\0&1&-1&0\\\end{pmatrix}$
$M^3=\begin{pmatrix}0&-1&1&0\\0&-1&0&1\\0&-1&0&0\\1&-1&0&0\\\end{pmatrix}$
$M^4=\begin{pmatrix}-1&1&0&0\\-1&0&1&0\\-1&0&0&1\\-1&0&0&0\\\end{pmatrix}$
$M^5=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}$
and our proposed cycle of coordinate points
$(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1),(-1,-1,-1,-1).$
As in the two-dimensional case, these points do not form a regular pentagon in the normal Cartesian space; but by using a modified vector norm (which again corresponds to not rendering the axes orthogonally), we gain the symmetry. The appropriate norm and its associated inner product are:
$||u||=||(u_1,u_2,u_3,u_4)||^2=u_1^2+u_2^2+u_3^2+u_4^2-\frac12(u_1u_2+u_1u_3+...+u_3u_4) =\frac1{16}(u_1+u_2+u_3+u_4)^2+\frac5{16}(u_1+u_2-u_3-u_4)^2+\frac5{16}(u_1-u_2-u_3+u_4)^2+\frac5{16}(u_1-u_2+u_3-u_4)^2$
$(u\cdot v)=u_1v_1+u_2v_2+u_3v_3+u_4v_4-\frac14(u_1v_2+u_1v_3+...u_3v_4)$
Here the cross-product terms run through all distinct permutations, including the distinction hetween $u$ and $v$ in the inner product. We note that as in the two-dimensional case, the net effect is to reduce the weight of the "one-signed" coordinate direction along which the point with all coordinates $-1$ lies.
With the correct norm and inner product above, the proposed coordinate vectors all satisfy
$||u||=1, (u\cdot v)=-\frac14.$
This means our five points are the vertices of a $4$-simplex centered at the origin, and any of twelve cycles of five edges embedded in this simplex will be a regular (equilateral, equiangular), nonplanar pentagon.