Symmetry of charge distribution for an infinite sheet of charges and invariances of $E$ If we consider the symmetries, we get that the electric field is a function of only the perpendicular distance away from the sheet, but when we do the actual calculation it is constant. Why does symmetry fail to predict this? If I go up more, then I see more charges.
 A: It’s not just symmetry that tells you the direction. You also have to use the fact that the direction of E is straight away from plus charges. 
For example, the B field has different directionality.
Similarly, the $1/r^2$ nature of the field needs to be used to show that E is constant here. Symmetry and/or geometry cannot by themselves be enough. 
For example, a $1/r^3$ force would drop off with distance from the plane, even though symmetry and geometry are the same for it. 
A: To add to Bob's answer, dimensional analysis (along with symmetry) is sufficient to tell you that the field must be independent of the perpendicular coordinate: the electric field must have dimensions of $[Q] [L]^{-2} [\epsilon_0]^{-1}$, since $$[E] = \left[\frac{q}{4\pi \epsilon_0 r^2}\right].$$
However, in this problem, we have two quantities with dimension, $\sigma$ (charge per unit area) and $z$ (the perpendicular distance). It should be obvious from this that $$E\propto \frac{\sigma}{\epsilon_0},$$ modulo some numerical factor.
Of course, the reason that $z$ doesn't come into play is because there is no other length-scale associated with the problem. Supposing, instead, that the plate was finite with some length $L$, then in general (on purely dimensional grounds) you could expect the electric field at to be some $$E \propto \frac{\sigma}{\epsilon_0} \left(a_0 + a_1 \left(\frac{z}{L}\right) + a_2 \left(\frac{z}{L}\right)^2 + ... \right) = \frac{\sigma}{\epsilon_0} f\left(\frac{z}{L}\right),$$ where $f$ is some function.
