Let $\tau=2\pi$ $=360^\circ$.
An $n$-gon antiprism has dihedral angles
$$\theta_n = \arccos\left(-\frac1{\sqrt3}\tan\frac\tau{4n}\right)$$
(where an $n$-gon meets a triangle) and
$$\phi_n = 2\arccos\left(\frac2{\sqrt3}\sin\frac\tau{4n}\right)$$ $$=\arccos\left(-\frac13\left(4\cos\frac\tau{2n}-1\right)\right)$$
(where two triangles meet). As $n$ increases from $3$ (the triangular antiprism being the Platonic octahedron), $\theta_n$ decreases from $\arccos({-\tfrac13})\approx109^\circ$ and approaches $\tfrac\tau4=90^\circ$, and $\phi_n$ increases from $\arccos({-\tfrac13})\approx109^\circ$ and approaches $\tfrac\tau2=180^\circ$.
Are there integers $l,m,n\geq3$ such that
$$\theta_l + \theta_m + \phi_n = \tau \; ?$$
If two polyhedra are required to have the same type of face where they touch, then we must have $l=m$. (And the other faces are all triangles, so there's no problem.) The equation becomes $2\theta_m + \phi_n = \tau$.
$$\theta_m=\frac\tau2-\frac{\phi_n}2$$ $$\cos\theta_m=-\cos\frac{\phi_n}2$$ $$-\frac1{\sqrt3}\tan\frac\tau{4m}=-\frac2{\sqrt3}\sin\frac\tau{4n}$$ $$\tan^2\frac\tau{4m}=4\sin^2\frac\tau{4n}$$ $$\frac{1-\cos\frac\tau{2m}}{1+\cos\frac\tau{2m}}=2\left(1-\cos\frac\tau{2n}\right)$$
This last equation can be solved for either cosine in terms of the other cosine. Thus $\mathbb Q(\cos\tfrac\tau{2m})=\mathbb Q(\cos\tfrac\tau{2n})$. But such fields are never equal unless $m=n$. ( Prove that $\mathbb Q(\cos\tfrac\pi7)\neq\mathbb Q(\cos\tfrac\pi9)$ ) Then the middle equation gives $\sec\tfrac\tau{4m}=2$, so $m=3/2$ which isn't an integer. (Anyway, the $3/2$ (crossed) antiprism is worse than non-convex; it's planar, with all faces coinciding.)
It remains to consider the case $l\neq m$.
$$\cos(\theta_l+\theta_m)=\cos(\tau-\phi_n)$$ $$\cos\theta_l\cos\theta_m-\sin\theta_l\sin\theta_m=\cos\phi_n$$ $$(\cos\theta_l\cos\theta_m-\cos\phi_n)^2=(1-\cos^2\theta_l)(1-\cos^2\theta_m)$$ $$-2\cos\theta_l\cos\theta_m\cos\phi_n+\cos^2\phi_n+\cos^2\theta_l+\cos^2\theta_m=1$$ $$\frac29\tan\frac\tau{4l}\tan\frac\tau{4m}\left(4\cos\frac\tau{2n}-1\right)+\frac13\tan^2\frac\tau{4l}+\frac13\tan^2\frac\tau{4m}+\frac19\left(4\cos\frac\tau{2n}-1\right)^2=1$$