Condition that a polygon is regular in the complex plane

Working in the complex plane throughout ...

Let $R_{n,k}$ be the "standard" regular $\{n/k\}$-gon (that is, the regular polygon with Schläfli symbol $\{n/k\}$ (where we leave the fraction un-reduced, and we allow $k=0$), whose vertices are at $\omega_n^{mk}$ for $m=0,1,\ldots,n-1$, and where $\omega_n$ is the principal $n$-th root of unity. (These vertices follow Grünbaum's convention, not depicted in Wikipedia, when $n$ and $k$ are not relatively prime. For instance, wikipedia shows $\{8/2\}$ as two squares overlapping at an angle; Grünbaum's convention considers $\{8/2\}$ to be a single $\{4/1\}$ square, traversed twice.) Here are the $\{5/k\}$-gons:

Both $R_{5,1}$ and $R_{5,4}$ are "regular convex pentagons" (and both $R_{5,2}$ and $R_{5,3}$ are "regular pentagrams"), but their opposing orientations are important here. Note that $R_{5,0}$'s vertices all coincide at $1+0i$, but there are still five of them.

For an $n$-gon $P$, let $[P]$ be the vector of its vertices (taken in order around the polygon) $$[P]:=(p_0, p_1, p_2,\ldots, p_{n-1})$$ As it happens, we can write uniquely $$[P] = a_0 [R_{n,0}] + a_1 [R_{n,1}] + \cdots + a_{n-1} [R_{n,n-1}] \tag{1}$$ for appropriate complex coefficients $a_k$. (Proof: Equating individual vector components gives a linear system of $n$ equations in the $n$ unknowns $a_k$.) Observe that each $a_k [R_{n,k}]$ is the vertex-vector of a scaled-and-rotated copy of $R_{n,k}$; this says that

Any $n$-gon is a "vertex sum" of $n$ regular $\{n/k\}$-gons (one for each $k$).

Because the centroid of polygons $R_{n,k}$ for $k\neq 0$ are at the origin, the centroid of $P$ (ie, the polygon's location in the plane) is determined by the $R_{n,0}$ term. On the other hand, the shape of $P$ is determined by those $k\neq 0$ terms, and we can say

$P$ is a regular $\{n/k\}$-gon, for $k\neq 0$, if and only if $a_k\neq 0$, and $a_m=0$ for $0<m\neq k$.

If we only care about non-oriented shapes, we can test whether $P$ is an $\{n/k\}$-gon or an $\{n/(n-k)\}$-gon; ie,

$P$ is a non-oriented regular $\{n/k\}$-gon, for $k\neq 0$, if and only if $a_m\neq0$ for $m=k$ or $m=n-k$ (but not both, unless $k=n/2$), and $a_m=0$ for all other $m>0$.

Now, OP's summation with roots of unity, call it $S_1(P)$, amounts to the dot product of a polygon's vertex-vector with that of $R_{n,1}$. Considering more-generally the dot product with vertex-vector of $R_{n,k}$, we have, invoking the decomposition $(1)$, $$S_k(P) := [P]\cdot [R_{n,k}] = \sum_{j=0}^{n-1} a_j [R_{n,j}]\cdot [R_{n,k}] \tag{2}$$ But $$[R_{n,j}]\cdot [R_{n,k}]=\sum_{m=0}^{n-1} \omega_n^{mj} \omega_n^{mk}=\sum_{m=0}^{n-1}\left(\omega_n^{j+k}\right)^m=\begin{cases} n, & j+k=0 \pmod{n} \\ 0, & \text{otherwise}\end{cases} \tag{3}$$ Therefore, $(2)$ reduces to $$S_k(P) = n a_{n-k} \tag{4}$$ (In the case of $k=0$, we take $a_{n-0}=a_0$.) Therefore, the test for (oriented) regularity becomes

$P$ is a regular $\{n/k\}$-gon if and only if $S_{n-k}(P)\neq 0$, and $S_m(P)=0$ for $0<m\neq n-k$.

For non-oriented regularity:

$P$ is a non-oriented regular $\{n/k\}$-gon if and only if $S_m(P)\neq0$ for $m=k$ or $m=n-k$ (but not both, unless $k=n/2$), and $S_m(P)=0$ for all other $m>0$.

In the case of a triangle $T$, decomposition $(1)$ is simply $$[T] = a_0[R_{3,0}] + a_1[R_{3,1}]+ a_2[R_{3,2}] \tag{1.3}$$ OP's test for regularity (that is, equilaterality) amounts to checking whether $S_1(T)=0$, which corresponds to whether the $R_{3,2}$ component of $(1)$ vanishes. This is consistent with the oriented test given above. For a non-oriented test, we'd check that $S_1(T)=0$ or $S_2(T)=0$ (but not both).

So, for general polygons, OP was correct to suspect a connection (congratulations on the insight!), but simply didn't allow for enough tests.

The ideas here are captured in a modified theorem of Barlotti. Barlotti observed that any $n$-gon is a "vertex sum" of affinely-regular non-oriented $\{n/k\}$-gons (with $0\leq k\leq n/2$). And it happens that, for $k\neq0$ and $k\neq n/2$, an affinely-regular $\{n/k\}$-gon is itself the vertex sum of oppositely-oriented actually-regular ones, so that we get a sum ($(1)$ above) with an $\{n/k\}$ component for each $0\leq k\leq n-1$.

(One can also write any polyhedron as a "vertex sum" of combinatorially-equivalent "symmetric" polyhedra. The nature of the "symmetric" components, and how many there are, depend upon eigen-analysis of the adjacency matrix of the polyhedron's vertex-and-edge skeleton. Likewise in higher dimensions. But I digress ...)

An interesting related result is a generalization of Napoleon's Theorem:

Given an oriented $n$-gon, erect an $\{n/k\}$-gon, for some $k$, on each edge, and connect their centroids in order to create a new oriented polygon. (Note that if $k=0$, the new polygon matches the old, so we could ignore it.) On the sides of that polygon, erect $\{n/k\}$-gons, for a different $k$; iterate.

After $n-1$ steps (or $n-2$ steps, if we're ignoring $k=0$), the resulting polygon will be a regular $\{n/k\}$-gon, for the leftover value of $k$!

As it happens, the order of the $k$s used in the iteration doesn't matter.

Napoleon's Theorem for an oriented triangle (ignoring $k=0$) erects $\{3/1\}$-triangles on the sides to get a regular $\{3/2\}$-triangle; and erects $\{3/2\}$-triangles on the sides to get a regular $\{3/1\}$-triangle. (The two results are the "inner" and "outer" Napoleon triangles. Which is which depends upon how the original triangle is oriented.) Likewise, the iterative process above associates with every $n$-gon a regular "leftover" $\{n/k\}$-gon for every $k$ ... kinda like the decomposition $(1)$. However, these Napoleonic polygons do not typically match the size and/or rotation of the components in $(1)$.