Find all integers $n\geq 3$ such that there exists a convex $n$-gon $A_1A_2\dots A_n$ which satisfies the following conditions:
 - All interior angles of the polygon are equal
 - Not all sides of the polygon are equal
 - There exists a triangle $T$ and a point $O$ inside the polygon such that the $n$ triangles $OA_1A_2,\ OA_2A_3,\  \dots,\ OA_{n-1}A_n,\ OA_nA_1$ are all similar to $T$, not necessarily in the same vertex order.
To solve the problem, we need to find all integers \( n \geq 3 \) such that there exists a convex \( n \)-gon \( A_1A_2 \dots A_n \) with the following properties:
1. All interior angles of the polygon are equal.
2. Not all sides of the polygon are equal.
3. There exists a triangle \( T \) and a point \( O \) inside the polygon such that the \( n \) triangles \( OA_1A_2, OA_2A_3, \dots, OA_{n-1}A_n, OA_nA_1 \) are all similar to \( T \), not necessarily in the same vertex order.

We will proceed by proving several claims and using them to determine the possible values of \( n \).

1. **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:
   \[
   \angle PA_iA_{i+1} = \angle PA_{i+1}A_i \implies PA_i = PA_{i+1}
   \]
   This implies that all the triangles \( \triangle PA_iA_{i+1} \) are congruent, so all the sides \( A_iA_{i+1} \) are equal, which contradicts the condition that not all sides are equal. 
   \(\blacksquare\)

2. **Claim 2: One of the angles of \( T \) is \( \frac{360^\circ}{n} \).**

   **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 \( \frac{360^\circ}{n} \), as desired.
   \(\blacksquare\)

   Let the angles of \( T \) be \( \gamma = \frac{360^\circ}{n} \), \( \alpha \), and \( \beta \).

3. **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, which is a 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 \( \gamma \leq \alpha, \beta \).

4. **Claim 4: We must have \( n \leq 6 \).**

   **Proof:**
   Suppose that \( n > 6 \). This means that \( \gamma = \frac{360^\circ}{n} < 60^\circ \), 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 than \( \gamma \), and the other is larger. Without loss of generality, 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. Without loss of generality, let \( \angle OA_iA_{i+1} = \gamma \). Now, note that:
   \[
   \angle OA_iA_{i+1} = 180^\circ - \gamma - \angle OA_iA_{i+1} = 180^\circ - 2\gamma = \alpha + \beta - \gamma
   \]
   Note that this cannot be equal to \( \alpha \) or \( \beta \), since \( \alpha < \gamma < \beta \). Therefore, we must have:
   \[
   \angle OA_iA_{i+1} = \gamma = 180^\circ - 2\gamma \implies \gamma = 60^\circ \implies n = 6
   \]
   This is a contradiction.
   \(\blacksquare\)

   From here, we eliminate the smaller cases:
   - For \( n = 3 \): An equiangular triangle must be an equilateral triangle.
   - For \( n = 4 \): Note that \( A_1A_2A_3A_4 \) is a rectangle, and \( \gamma = 90^\circ \). However, these \( \gamma \)'s cannot appear on the edge of the polygon (i.e., \( \angle OA_iA_{i\pm 1} = 90^\circ \)), since the interior angle is also \( 90^\circ \). This means \( \angle A_iOA_{i+1} = \gamma \) for all \( i \), contradicting Claim 3.
   - For \( n = 5 \): Note that \( \gamma = 72^\circ \). Without loss of generality, from Claim 3, suppose we have \( \angle OA_1A_2 = 72^\circ \), meaning that \( \angle OA_1A_4 = 36^\circ \). Therefore, we have that \( \{\alpha, \beta\} = \{36^\circ, 72^\circ\} \). However, the only way to combine 5 of these angles to get a total of \( 360^\circ \) in the center is to use five \( 72^\circ \) angles, contradicting Claim 3.

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

The final answer is \( \boxed{ n = 6 } \).