In $7$-gon $A_1A_2A_3A_4A_5A_6A_7$ diagonals $A_1A_3, A_2A_4, A_3A_5, A_4A_6, A_5A_7, A_6A_1$ and $A_7A_2$ are congruent to each other and diagonals $A_1A_4, A_2A_5, A_3A_6, A_4A_7, A_5A_1, A_6A_2$ and $A_7A_3$ are also congruent to each other. Is the polygon necessarily regular?
1. **Identify the given conditions:**
   - We have a 7-gon \( A_1A_2A_3A_4A_5A_6A_7 \).
   - Diagonals \( A_1A_3, A_2A_4, A_3A_5, A_4A_6, A_5A_7, A_6A_1, \) and \( A_7A_2 \) are congruent.
   - Diagonals \( A_1A_4, A_2A_5, A_3A_6, A_4A_7, A_5A_1, A_6A_2, \) and \( A_7A_3 \) are also congruent.

2. **Analyze the triangles formed by the diagonals:**
   - Consider the triangles \( \triangle A_{i-2}A_iA_{i+2} \). These triangles are formed by the diagonals \( A_{i-2}A_i \) and \( A_iA_{i+2} \), which are congruent by the given conditions.
   - Since these diagonals are congruent, the triangles \( \triangle A_{i-2}A_iA_{i+2} \) are all congruent to each other.

3. **Calculate the angles:**
   - In a regular 7-gon, the internal angle at each vertex is \( \frac{5\pi}{7} \).
   - The angle \( \angle A_{i-2}A_iA_{i+2} \) is formed by skipping one vertex, so it is \( \frac{3\pi}{7} \).

4. **Prove the heptagon is cyclic:**
   - Since all the triangles \( \triangle A_{i-2}A_iA_{i+2} \) are congruent and the angles \( \angle A_{i-2}A_iA_{i+2} \) are equal, the heptagon must be cyclic. This means that all vertices lie on a common circle.

5. **Conclude the regularity of the heptagon:**
   - A cyclic heptagon with congruent diagonals as given must be regular. This is because the equal diagonals and equal angles imply that all sides are also equal.

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\boxed{\text{True}}
\]