If two lines are perpendicular to the base, they are parellel with each other, they won't intersect, therefore not forming a pyramid. So the maximum is $1$. Q.E.D.

Domingues3 wrote: If two lines are perpendicular to the base, they are parellel with each other, they won't intersect, therefore not forming a pyramid. So the maximum is $1$. Q.E.D. They can intersect). If they are perpendicular to the base that doesn't mean that they can't intersect

Note that it's not hard to see but here I'll write it clearly, so it may look a bit long. I will assume that the $n$-gon don't have to be a convex polygon since it didn't require in the problem. WLOG that the polygon lie on the plane $z=0$, denote their vertices consecutively by $(x_1,y_1,0),(x_2,y_2,0),...,(x_n,y_n,0)$. Then suppose the vertices of the pyramid have coordinate $(p,q,r)$. Note that a triangle formed by three edges connected $(p,q,r),(x_i,y_i,0),(x_j,y_j,0)$ is a right triangle (with one of it's legs is $(x_i,y_i,0),(x_j,y_j,0)$) iff the triangle formed by three edges connected $(p,q,0),(x_i,y_i,0),(x_j,y_j,0)$ is also a right triangle. So we can assume that $r=0$ and then the problem reduce to two dimensions. Start with two points $(p,q),(x_1,y_1)$ It's clear that we can inductively (in clockwise angle) construct $(x_i,y_i)$ so that the angle formed by $(p,q),(x_i,y_i)$ and $(x_i,y_i),(x_{i-1},y_{i-1})$ for each $i=2,3,...,n$ Then we draw the $n^{th}$ line to complete the (possibly concave) polygon. Then we've to show that there not exist such figure with $n$ perpendicular edges. Observe that the right angle doesn't occur at the same vertex twice (otherwise, two sides of the polygon is the same line.). By Pythagoras, we have $(p-x_i)^2+(q-x_i)^2)-((p-x_{i+1})^2+(q-y_{i+1})^2)=(x_i-x_{i+1})^2+(y_i-y_{i+1})^2$ This must hold for all $i=1,2,...,n$ (indices taken modulo $n$). Summing up all equalities, we get a contradiction since $LHS=0,RHS>0$.

Domingues3 wrote: If two lines are perpendicular to the base, they are parellel with each other, they won't intersect, therefore not forming a pyramid. So the maximum is $1$. Q.E.D. just say : they are parallel but theyhave a common point so they coincide. RH HAS