I proposed this problem : D which also indirectly killed geo in the test D: I think this turned out pretty nicely and is somewhat of a surprising result, but the first rough drafts of the problem were pretty horrible (for example, $T$ wasn't fixed and instead was allowed to vary in the sets of isosceles or right triangles depending on each $i$). I'm glad it ended up coalescing into something reasonable.

The only possible value of $n$ is $\boxed{n=6}$. We first assert some claims (throughout the solution, indices are assumed to be modulo $n$): Claim 1: The interior angle bisectors of $A_1A_2A_3\dots A_n$ cannot concur at a point. Proof: Since the polygon is equiangular, if all of these angle bisectors concurred at $P$, then we would have $\angle PA_iA_{i+1}=\angle PA_{i+1}A_i\implies PA_i=PA_{i+1}$, meaning that all of the triangles $\triangle PA_iA_{i+1}$ are congruent, so all the sides $A_iA_{i+1}$ are equal, contradiction. $\blacksquare$ Claim 2: One of the angles of $T$ is $360/n^{\circ}$. Proof: From Claim 1, we note that for some $i$, we must have $\angle OA_iA_{i-1}\neq \angle OA_iA_{i+1}$, meaning that these two angles must be angles in $T$. Therefore, the last angle is the exterior angle of the original polygon, which is $360/n^{\circ}$, as desired. $\blacksquare$ Let the angles of $T$ be $\gamma=360/n^{\circ}$, $\alpha$, and $\beta$. Claim 3: We cannot have $\angle A_iOA_{i+1}=\gamma$ for all $i$. Proof: Suppose this were the case. Then, we would have $$\angle OA_1A_2=180^{\circ}-\gamma-\angle OA_2A_1=180^{\circ}-\gamma-(180^{\circ}-\gamma-\angle OA_2A_3)=\angle OA_2A_3.$$Continuing in this line, we would have $\angle OA_1A_2=\angle OA_2A_3=\angle OA_3A_4=\dots =\angle OA_nA_1$. Combining this with the original assumption gives that $\triangle OA_1A_2\sim \triangle OA_2A_3\sim \dots\sim \triangle OA_nA_1$, where the orientation is in vertex order. However, this yields: \[1=\frac{OA_2}{OA_1}\cdot\frac{OA_3}{OA_2}\cdot \dots \cdot \frac{OA_1}{OA_n}=\left(\frac{OA_2}{OA_1}\right)^n\implies OA_1=OA_2\implies OA_1=OA_2=OA_3=\dots=OA_n.\]This means that the similar triangles are actually congruent, so the sides are all equal, contradiction. $\blacksquare$ This means that we must replace some of the $\gamma$'s in the center ($\angle A_iOA_{i+1}$'s) with $\alpha$'s and $\beta$'s. However, since all $n$ of the $\gamma$'s snugly fit in $360^{\circ}$ of space, we cannot replace the $\gamma$'s with anything bigger (without adding smaller angles). This means that we cannot have $\gamma\leq \alpha, \beta$. We now introduce the killer claim: Claim 4: We must have $n\leq 6$. Proof: Suppose that $n>6$. This means that $\gamma=360/n < 60$, meaning that we cannot have $\gamma \geq \alpha, \beta$ (since they all sum to $180^{\circ}$). Combining this with the previous inequality gives that one of $\alpha,\beta$ is smaller that $\gamma$, and the other is larger. WLOG, suppose we have $\alpha<\gamma<\beta$. From Claim 3, we cannot have all of the $\gamma$'s in the center, so at least one $\gamma$-measured angle is on the perimeter. WLOG, let $\angle OA_iA_{i+1}=\gamma$. Now, note that $\angle OA_iA_{i+1}=180-\gamma-\angle OA_iA_{i+1}=180-2\gamma=\alpha+\beta-\gamma$. Note that this cannot be equal to $\alpha$ or $\beta$, since $\alpha<\gamma<\beta$. Therefore, we must have $\angle OA_iA_{i+1}=\gamma=180-2\gamma\implies \gamma=60\implies n=6$, contradiction. $\blacksquare$ From here, we eliminate the smaller cases:

n=3

n=4

n=5

A family of constructions for $n=6$ is listed below, for $\alpha+\beta=120^{\circ}$ and $\alpha\neq \beta$ (verification is not difficult with Law of Sines): \[(OA_1,OA_2,OA_3,OA_4,OA_5,OA_6)=\left(\sin\alpha, \frac{2}{\sqrt{3}}\sin^2\alpha, \sin\alpha, \sin\beta,\frac{2}{\sqrt{3}}\sin^2\beta, \sin\beta\right)\]\[(\angle A_1OA_2, \angle A_2OA_3, \angle A_3OA_4, \angle A_4OA_5, \angle A_5OA_6, \angle A_6OA_1)=(\beta, \beta, 60^{\circ}, \alpha, \alpha, 60^{\circ})\]