Convex pentagons are similar if conformally equivalent. The problem:
Suppose two convex pentagons $A$ and $B$ have equal interior angles (that is, $A=A_1A_2A_3A_4A_5$ and $B=B_1B_2B_3B_4B_5$) with $\angle A_j =\angle B_j$ for each $j\in\{1,\ldots,5\}$). 
Suppose that $\mbox{int}(A) \approx \mbox{int}(B)$ are conformally equivalent with a biholomorphism $f:\mbox{int}(A) \rightarrow \mbox{int}(B)$ whose continuous extension to the boundary maps $A_i\overset{f}{\mapsto}B_i$. 
Show that under these conditions, $A$ and $B$ are similar. 
Ideas:
I would suspect the reflection principle would be applicable, but I'm not certain how to work out the proof. 
 A: Let $f_1,f_2$ be conformal maps of the upper half-plane onto your polygons, and $f$ is your map between the polygons. Then $g:=f_1^{-1}\circ f\circ f_2$
is a conformal automorphism of the upper half-plane. Now consider the functions
$f_2$ and $f_1\circ g$. Both map the upper half-plane to polygons, and there are $5$ points $a_1,\ldots,a_5$ on the real line such that for each $k$, both
$f_2$ and $f_1\circ g$ map $a_k$ to vertices of their respective polygons, and the interior angles at these two vertices are the same angle.
Now both $f_2$ and $f_1\circ g$ must be represented by the Schwarz-Christoffel formula with the same singularities and same angles. But angles and and singularities determine the Schwarz--Christoffel formula up to a composition with an affine map. Therefore
$$f_2=Af_1\circ g+B$$
which proves the statement.
Remarks. 5 is irrelevant. Convexity is also irrelevant. All we need is that 
the conformal map $f$ between the polygons sends vertices to vertices and the interior angles at the corresponding vertices are equal.  
