Call $n$ good if the regular $n$-gon can be triangulated with isosceles triangles. By segments we mean the sides and the diagonals of the $n$-gon; the sides are the shortest among all segments.

Let $n$ be good and $T$ an isosceles triangulation of the regular $n$-gon $P$. Suppose that the base of a triangle $\Delta \in T$ is a side $a$ of $P$. Then the vertex of $\Delta$ opposing $a$ is on the perpendicular bisector of $a$, which passes through the center of $P$, and also the circumcircle of $P$. Hence $n$ is odd and the center of $P$ is interior to $\Delta$, implying that such a triangle $\Delta$ is unique.

Let $n$ be even. Then sides are the shortest segments and none of them is a base of a triangle of $T$. So all of them are divided into pairs of consecutive ones, and each pair contains the equal sides of a triangle of $T$. Deleting these $\frac{n}{2}$ isosceles triangles leaves a regular $\frac{n}{2}$-gon which therefore also admits of an isosceles triangulation. It follows that an even $n \ge 6$ is good if and only if so is $\frac{n}{2}$.

Let $n$ be odd. Then the sides cannot be paired up like in the even case, one of them must be a base of a triangle $\Delta$ from $T$ as explained earlier. The equal sides of $\Delta$ are diagonals of $P$ (longest ones). Removing $\Delta$ leaves two congruent polygons which must have isosceles triangulations. Let $P_1$ be one of them. It has $n_1 = \frac{1}{2}(n(n+1))$ sides; thus $n_1$ is good. One of the sides is a diagonal $d_1$, the rest are sides of $P$, hence shorter. So $d_1$ is a base of a triangle $\Delta_1$ of $T$, and its opposite vertex divides the remaining $n_1 - 3$ vertices into two equal halves. It follows that $n_1$ is odd. Remove $\Delta_1$ from $P_1$ and denote by $P_2$ one of the two obtained congruent polygons with $n_2 = \frac{1}{2}(n_4 + 1)$ sides; $n_2$ is good. The same argument applies to $P_2$ because one of its sides is a diagonal $d_2$, and the rest are sides of $P$. We conclude that $n_2$ is odd then define $n_3 = \frac{1}{2}(n_2 + 1)$, and so on. Thus each of the numbers $n > n_1 > n_2 > \dots$ is odd and good, as long as it is $\ge 3$. Let $k$ be such that $n_k \ge 3 > n_{k+1}$. If $n_{k+1} = 1$ then $n_k = 1$ which is false. So $n_{k+1} = 2$ and so $n_k = 3$. Write $n_k = 3 = 2^1 + 1$ and backwards to obtain $n_{k-1} = 2^2 + 1$ and likewise $n_{k-2} = 2^3 + 1$, ..., $n_1 = 2^k + 1$, $n = 2^{k+1} + 1$. Therefore $n-1$ is a power of 2. In addition the steps of the argument imply a construction showing that the converse is also true.

To sum up, consider two cases for a general $n$. If $n \ge 4$ is a power of 2, $n = 2^m$ with $m \ge 2$, then it is good if and only if so are $2^{m-1}, 2^{m-2}, \dots, 2^2 = 4$. Since the square has an isosceles triangulation, the powers of 2 are good. If $n \ge 3$ is not a power of 2 then $n = 2^m$ with $k \ge 3$ odd and $m \ge 0$. By the above, $n$ is good if and only if so is $k$, and the latter holds if and only if $k$ is of the form $k = 2^l + 1$ with $l \ge 1$. Hence $n = 2^n + 2^l$ with $u > v \ge 0$. In conclusion the good numbers are $2^m$ with $m \ge 2$ and $2^n + 2^l$ with $u > v \ge 0$.