CroatianCompetitions

Let the new polygon have $m$ vertices. Two cases are possible:

Case 1. The new polygon lies outside the given polygon. In other words, $m$-gon and $n$-gon are on opposite sides of the line $A_2A_3$.
In this case $\angle BA_2A_3 + \angle A_1A_2A_3 + 60^\circ = 360^\circ$.
$$\frac{n-2}{n} \cdot 180^\circ + \frac{m-2}{m} \cdot 180^\circ + 60^\circ = 360^\circ$$
$$3m(n-2) + 3n(m-2) + mn = 6mn$$
$$mn - 6m = 6n$$
$$m = \frac{6n}{n-6} = 6 + \frac{36}{n-6}.$$
Obviously $n-6 \in \mathbb{N}$ and $n-6$ divides $36$.
Checking all the possibilities we find nine solutions:

Case 2. The observed $m$-gon and $n$-gon are on the same side of the line $A_2A_3$.
In this case we have $\angle BA_2A_3 = \angle A_1A_2A_3 + 60^\circ$
$$\begin{aligned} \frac{m-2}{m} \cdot 180^\circ &= 60^\circ + \frac{n-2}{n} \cdot 180^\circ \\ 3n(m-2) &= mn + 3m(n-2) \\ 6n + nm &= 6m \\ n &= \frac{6m}{m+6} = 6 - \frac{36}{m+6}. \end{aligned}$$
Since $n \ge 3$, $\frac{36}{m+6}$ must belong to $\{1, 2, 3\}$. From there we get another three solutions
$$\begin{array}{c|ccc} m+6 & 36 & 18 & 12 \\ m & 30 & 12 & 6 \\ n & 5 & 4 & 3 \end{array}$$