If we identify the 4 sides of a regular tetrahedron in $\mathbb{R}^3$ by letting the group of all isometries of the tetrahedron act on it, what would the resulting space look like?
The resulting space will be just a triangle with isotropy groups $\mathbb{Z}/3$ for points on its sides and the dihedral group of order 8 for the 3 vertices?
From this picture you will see that the resulting quotient orbifold is a triangle with isotropy groups $\mathbb{Z}/2$ on each edge, with the order $6$ dihedral group on two of the vertices, and the order $4$ dihedral group on the remaining vertex.