Denote such polygon by $P$, let $P’$ be the convex hull of all $n$ vertices of $P$. Suppose the vertices of $P’$ are $W_1,W_2,...,W_{n’}$. If there is at least one vertex of $P$ that lie inside $P’$, there exists $i\in \{ 1,2,...,n’\}$ such that $W_iW_{i+1}$ is not a segment joining two neighboring vertices of $P$ (here, we assume $W_{n’+1}=W_1$). Then, for any other vertex $W_j$ of $P’$, which is a vertex of $P$, the reflection of $W_j$ over $W_iW_{i+1}$ will fall outside of $P’$ (and also outside of $P$), contradiction. So all vertices of $P$ are contained in $P’$, which means $P$ must be convex. Suppose vertices of $P$ are $V_1,V_2,...,V_n$, we denote $\angle{V_i}=\angle{V_{i-1}V_iV_{i+1}}$. For each $i\in \{ 1,2,...,n\}$ (all indices are considered in modulo $n$), since the reflection of $V_{i+2}$ and $V_{i-2}$ over diagonal $V_{i-1}V_{i+1}$ must fall not outside of $P$, we get that $\angle{V_iV_{i-1}V_{i+1}} \geq \angle{V_{i+1}V_{i-1}V_{i-2}}$ and $\angle{V_iV_{i+1}V_{i-1}} \geq \angle{V_{i-1}V_{i+1}V_{i+2}}$. This give us $\angle{V_{i-1}}\leq 2\angle{V_iV_{i-1}V_{i+1}}$ and $\angle{V_{i+1}} \leq 2\angle{V_iV_{i+1}V_{i-1}}$. So $\angle{V_{i-1}}+2\angle{V_i}+\angle{V_{i+1}}\leq 360^{\circ}$. Summing the inequalities for all $i$ gives us $4\times 180(n-2)\leq 360n\Rightarrow n\leq 4$. But since $n\geq 4$, all equalities must hold. Hence $n=4$ and we get $4$ pairs of equal angles. This easily give all possible such polygon are rhombus.