 Solution

Let $r_1,r_2$ be the internal angle of refraction values measured from the two face normals, and let $i,e$ be the external incidence and emergence angles. Geometry of an optical prism gives $r_1+r_2=A$ and total deviation $\delta =(i-r_1)+(e-r_2)=i+e-A$.

For an isosceles optical prism whose internal beam is parallel to its base, reflection symmetry gives $r_1=r_2=A/2$. Applying Snell's law at both faces, with air refractive index approximated by one, gives $\sin i=n\sin (A/2)=\sin e$. For transmitted forward rays, $i=e$, so $i=(\delta +A)/2$. HenceThis is the symmetric minimum deviation geometry at a single wavelength. Transmission requires $n\sin (A/2)\le 1$; otherwise the proposed symmetric path cannot enter and leave through the two faces.