$ABC$ is a triangle. Reflect each vertex in the opposite side to get the triangle $A'B'C'$. Find a necessary and sufficient condition on $ABC$ for $A'B'C'$ to be equilateral.
1. **Reflecting Vertices and Congruent Triangles**:
   - Reflect each vertex of $\triangle ABC$ across the opposite side to get $\triangle A'B'C'$.
   - By reflection properties, $\triangle BAC \cong \triangle B'AC$ and similarly for the other vertices.

2. **Angle Calculation**:
   - Since $\triangle BAC \cong \triangle B'AC$, we have $\angle B'AC' = 3\angle A$.
   - Using the cosine rule, we can express the sides of $\triangle A'B'C'$:
     \[
     B'C'^2 = b^2 + c^2 - 2bc \cos 3A
     \]

3. **Equilateral Condition**:
   - For $\triangle A'B'C'$ to be equilateral, we need:
     \[
     b^2 + c^2 - 2bc \cos 3A = c^2 + a^2 - 2ca \cos 3B = a^2 + b^2 - 2ab \cos 3C
     \]
   - This simplifies to:
     \[
     a^2 + 2bc \cos 3A = b^2 + 2ca \cos 3B = c^2 + 2ab \cos 3C
     \]

4. **Cosine of Triple Angles**:
   - Using the triple angle formula for cosine:
     \[
     \cos 3A = 4\cos^3 A - 3\cos A
     \]
   - Substitute $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$:
     \[
     \cos 3A = \left(\frac{b^2 + c^2 - a^2}{2bc}\right)(1 - 4\sin^2 A)
     \]
   - Let $\sin^2 A = x$, $\sin^2 B = y$, $\sin^2 C = z$, and $R$ be the circumradius. By the sine rule:
     \[
     a^2 = 4R^2 x, \quad b^2 = 4R^2 y, \quad c^2 = 4R^2 z
     \]

5. **Simplifying the Equations**:
   - Substitute these into the equilateral condition:
     \[
     a^2 + 2bc \cos 3A = 4R^2 \left( x + (y + z - x)(1 - 4x) \right)
     \]
   - This gives us:
     \[
     x + (y + z - x)(1 - 4x) = y + (z + x - y)(1 - 4y) = z + (x + y - z)(1 - 4z)
     \]

6. **Solving the System**:
   - Simplify the equations:
     \[
     (x - y)(1 + 4z - 4x - 4y) = 0
     \]
   - If at least two of $(x - y), (y - z), (z - x)$ are zero, then $\triangle ABC$ must be equilateral.
   - If none are zero, we have the linear system:
     \[
     1 + 4z - 4x - 4y = 1 + 4x - 4y - 4z = 1 + 4y - 4z - 4x = 0
     \]
   - This system implies $x = y = z$, which is a contradiction unless $\triangle ABC$ is equilateral.

7. **Special Cases**:
   - If exactly one of $(x - y), (y - z), (z - x)$ is zero, assume $x = y$. Then:
     \[
     1 + 4x - 4y - 4z = 1 + 4y - 4z - 4x = 0
     \]
   - This implies $z = \frac{1}{4} = \sin^2 C \implies \sin C = \frac{1}{2} \implies \angle C = 30^\circ \text{ or } 150^\circ$.
   - Also, $x = y \implies \angle A = \angle B$.

Conclusion:
\[
\boxed{\triangle A'B'C' \text{ is equilateral} \iff \triangle ABC \text{ is isosceles with the angle between equal sides } 30^\circ, 60^\circ, \text{ or } 150^\circ}
\]