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 $\boxed{ n = 6 }$.