Solution. The answer is $n = 2^{m+1} + 2^k$, where $m$ and $k$ are nonnegative integers. In other words, $n$ is either a power of 2 (when $m+1=k$) or the sum of two nonequal powers of 2 (with $1=2^0$ being considered as a power of 2).
We start with the following observation.
Lemma. Let $Q = Q_0Q_1\dots Q_t$ be a convex polygon with $Q_0Q_1 = Q_1Q_2 = \dots = Q_{t-1}Q_t$. Suppose that $Q$ is cyclic and its circumcenter does not lie in its interior. If there is a triangulation of $Q$ consisting only of isosceles triangles, then $t = 2^a$, where $a$ is a positive integer.
Proof. We call an arc minor if its arc measure is less than or equal to 180°. By the given conditions, points $Q_1, \dots, Q_{t-1}$ lie on the minor arc $\text{Q 0Q t}$ of the circumcircle, so none of the angles $Q_iQ_jQ_k$ ($0 \le i < j < k \le t$) is acute. (See the left-hand side diagram shown below.) It is not difficult to see that $Q_0Q_t$ is longer than each other side or diagonal of $Q$. Thus $Q_0Q_t$ must be the base of an isosceles triangle in the triangulation of $Q$. Therefore, $t$ must be even. We write $t = 2s$. Then $Q_0Q_sQ_t$ is an isosceles triangle in the triangulation. We can apply the same process to polygon $Q_0Q_1\dots Q_s$ and show that $s$ is even. Repeating this process leads to the conclusion that $t = 2^a$ for some positive integer $a$.
The results of the lemma can be generalized by allowing $a=0$ if we consider the degenerate case $Q = Q_0Q_1$. $\square$

We are ready to prove our main result. Let $\mathcal{P} = P_1P_2...P_n$ denote the regular polygon. There is an isosceles triangle in the triangulation such that the center of $\mathcal{P}$ lies within the boundary of the triangle. Without loss of generality, we may assume that $P_1P_iP_j$, with $P_1P_i = P_1P_j$ (that is, $P_j = P_{n-i+2}$), is this triangle. Applying the Lemma to the polygons $P_1...P_i, P_i...P_j$, and $P_j...P_1$, we conclude that there are $2^m-1, 2^k-1, 2^m-1$ (where $m$ and $k$ are nonnegative integers) vertices in the interiors of the minor arcs $\widehat{P_1P_i}, \widehat{P_iP_j}, \widehat{P_jP_1}$, respectively. (In other words, $i = 2^m+1, j = 2^k+i$.) Hence
$$n = 2^m - 1 + 2^k - 1 + 2^m - 1 + 3 = 2^{m+1} + 2^k,$$
where $m$ and $k$ are nonnegative integers. The above discussion can easily lead to a triangulation consisting of only isosceles triangles for $n = 2^{m+1}+2^k$. (The middle diagram shown above illustrates the case $n = 18 = 2^{3+1}+2^1$. The right-hand side diagram shown above illustrates the case $n = 16 = 2^{2+1}+2^3$.)