A polygon with the same edge directions as a regular polygon The following is certainly not hard to prove, but I am specifically looking for a short and nice, but still rigorous argument. Please don't be hand-wavy.
Suppose $P\subset\Bbb R^2$ is a polygon with the following properties:


*

*$P$ is centrally symmetric,

*$P$ has all vertices on a common circle, and

*$P$ has the same edge-directions as some fixed regular $n$-gon.


In want to show that then $P$ is vertex-transitive, which is a fancy way to say that it is either regular, or has exactly two different edge length which alternate (see the picture below).


Consider  the following picture to see that central symmetry is necessary.

In particular, any proof that uses that we have


*

*equal interior angles, or

*exactly $n$ vertices


needs to apply central symmetry somewhere.
 A: Let's look at four consecutive vertices $P_1P_2P_3P_4$. The polygons has equal
angles by your third condition, so $\angle P_1 P_2 P_3=\angle P_2P_3P_4$.
Then these two angles subtend equal chords in the circle, so $|P_1P_3|=|P_2P_4|$.
Then the triangles $P_1P_2P_3$ and $P_4P_3P_2$ are congruent. So $|P_1P_2|=|P_3P_4|$: alternate sides of the polygon are equal.
A: The edges of the regular $n$-gon have either $n$ or $\frac{n}{2}$ different edge directions. By central symmetry, if $P$ is an $m$-gon then it has precisely $\frac{m}{2}$ different edge directions. Therefore either $m=2n$ or $m=n$.
Going around the polygon in an anticlockwise direction, each exterior angle is of the form $k_i\frac{2\pi}{n}$ where $k_i$ is a positive integer. Since $\sum k_i\frac{2\pi}{n}=2\pi$ we have each $k_i=1$ and $m=n$.
The two edges adjacent to any particular edge $e$ of the polygon therefore make equal angles with $e$ and therefore, by symmetry about the perpendicular bisector of $e$, they are of equal length.
A: Here's a couple of counterexamples (if we take "polygon" to mean "a closed path of line segments"):


Each diagram shows a "half-polygon" path from $P_0$ to its antipode, $P_5$, with edges parallel to those of a regular $10$-gon. (In fact, the vertices are those of the $10$-gon.) The complete polygon consists of this path and its image under a half-turn; the vertices are labeled in the appropriate order, but the edges are suppressed because some of them overlap.
The corresponding polygons are not vertex-transitive. That said, these polygons are a bit pathological by being self-intersecting and having coincident vertices and edges. Perhaps these are meant to be excluded.

Here's a fuller discussion ...
(Working on an answer to A polygon with the same edge directions as a regular polygon.)
Recapping my thoughts in the comments to the question:


*

*By central symmetry, we have an even number of vertices; say, $2n$. Now, although the order of the edge directions of property $(3)$ need not be consecutive, central symmetry tells us that $\overrightarrow{P_kP_{k+1}}=−\overrightarrow{P_{n+k}P_{n+k+1}}$; that is, the second half of the polygon uses the reverses of the edge directions from the first half, in the same order. Consequently, all we get to do is pick one edge direction from each parallel pair in a regular $2n$-gon to make a path between two diametrically-opposed points; central symmetry (in other words, rotating this path by a half-turn) completes the polygon.

*From any vertex, of the two choices for opposing edge directions to the next vertex, at least one direction leads out of the circle. (If they are tangent at the vertex, then both directions do this.) We can choose this "outward" direction if we make the corresponding edge length zero; but this is the same as using the "inward" direction (or either tangent), so we might as well say we chose that instead. As a result, our path is determined by a starting point and a permutation of the $2n$-gon's $n$ "unsigned" edge directions, and the need to get to the start's diametric opposite in $n$ steps.
Now, taking our regular $2n$-gon to have a "horizontal" edge, its $n$ "unsigned" edge directions are $\operatorname{cis}(k\pi/n)$, for $k=0,1,\ldots,n-1$, where I'm abusing notation to define $\operatorname{cis}\theta :=(\cos\theta,\sin\theta)$.
Given a vertex point $\operatorname{cis}\theta$ on the circle, and a direction $\operatorname{cis}\phi$, the "next" vertex is readily shown to be $(-\cos(\theta-2\phi), \sin(\theta-2\phi))$. Consequently, if $(p_0,p_1,p_2,\ldots,p_{n-1})$ is some permutation of $(0,1,2,\ldots,n-1)$, given a starting point $V_0$, we find that vertex $V_k$ is given by
$$V_k = \left((-1)^k \cos(\theta - \frac{2\pi}{n}\sigma_k), \sin(\theta-\frac{2\pi}{n}\sigma_k) \right), \qquad \sigma_k := \sum_{j=0}^{k-1}(-1)^j p_j$$
We want the end of our $n$-step path to meet the antipode of the starting point: $V_0 = -V_{n}$, so that $V_0\cdot V_{n}=-1$. 


*

*For $n$ odd, $V_0\cdot V_{n} = -\cos 2(\theta-\frac{\pi}{n}\sigma_{n})$. Setting this equal to $-1$ tells us that our starting angle satisfies, for some integer $m$,
$$2\left(\theta-\frac{\pi}{n}\sigma_{n-1}\right) = 2\pi m \quad\to\quad
\theta = \pi m + \frac{\pi}{n}\sigma_{n-1}$$ Since integer multiples of $\pi$ will simply give the path that number of half-turns, we can take $m=0$. Since $\theta$ is then a multiple of $\pi/n$, we have that $V_0$ (and all the vertices of our path and polygon) coincide with vertices of the regular $2n$-gon, as we saw in the "counterexample" $10$-gons above. In any case, for $n$ even, the resulting polygon is uniquely determined by the permutation of edge directions. 
However, multiple permutations give rise to congruent polygons: for instance, for the $10$-gon, considering permutations that start with $0$, the $24$ permutations of $(0,1,2,3,4)$ yield only the three different polygons: the regular $10$-gon itself (twice), and those indicated by the "counterexamples" (ten times each), and the "starry" $\{10/3\}$-gon.   

*For $n$ even, $V_0 \cdot V_{n} = \cos\left(\frac{2\pi}{n}\sigma_{n}\right)$; as this is independent of $\theta$, any starting point will work, provided that the permuted edge directions alone satisfy
$$\frac{2\pi}{n}\sigma_{n} = \pi + 2m\pi \quad\to\quad \sigma_{n} = \frac{n}{2}(2m+1)$$
The condition on the alternating sum excludes many permutations. For $12$-gons, of the $120$ zero-leading permutations of $(0,1,2,3,4,5)$, only $48$ satisfy the condition. Counting non-congruent instances is tricky, however, since the freedom to choose a starting point leads to infinite families of figures. When the starting vertex is at angle, say, $5\pi/12$, there are ten distinct polygons:


The question has asked whether such polygons are vertex-transitive. Clearly, they "usually" aren't. OP has commented an interest in just the convex figures; since the coordinate analysis can't see convexity, separate logic is needed.


*

*Recall that, in the case of a $2n$-gon for $n$ odd, the vertices coincide with those of the regular $2n$-gon. We can be more specific: since the path from starting vertex to antipode must use each unsigned direction once, it must use $n$ steps through $n+1$ vertices. Moreover, the path can't skip vertices without the full polygon having to cross itself to get back to them. Therefore, 



The only convex $2n$-gon, with $n$ odd, satisfying properties $(1)$, $(2)$, $(3)$ is the regular $2n$-gon itself, which is vertex-transitive.



*

*For $n$ even, things aren't quite so obvious, due to the freedom to choose the starting point. I suspect that convexity requires taking the unsigned edge directions in order, with appropriate restrictions placed on the starting point to avoid self-intersections, but I can't yet justify this.


It's worth noting that polygons can be non-convex yet vertex-transitive. See, for instance, the $12$-gon in row $2$, column $3$ of the figure, or consider the "starry" $\{10/3\}$-gon. (In general, a $\{2n/k\}$-gon with $k$ relatively prime to $2n$ is one of these figures.) So, just checking for edges that alternate between two lengths isn't quite enough to get at the specific figures in question.
This is about as far as I can take this analysis for the time being.
