= Solution
The boundary arcs meet at $0$ and $1$. The <Möbius transformation>
$$
w=\frac{z}{1-z}
$$
sends these points to $0$ and infinity, sends the real boundary segment to the positive real axis, and sends the circular boundary arc to the ray of argument $-\pi/6$. Hence it maps $D\cap L$ onto the sector
$$
-\frac{\pi}{6}<\arg w<0.
$$
The power $\zeta=w^6$ maps this sector conformally onto the lower half-plane. Finally, the Cayley map $\zeta\mapsto(\zeta+i)/(\zeta-i)$ maps the lower half-plane onto the unit disc. Thus a requested <conformal equivalence> is
$$
\boxed{F(z)=
\frac{\left(\dfrac{z}{1-z}\right)^6+i}
{\left(\dfrac{z}{1-z}\right)^6-i}}.
$$
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