5 constraints requires at least 5 coefficients, which would mean a quartic. But I'm guessing that you want rotational symmetry, f(1/2+x)-1/2 = 1/2-f(1/2-x), which requires a polynomial of odd degree, so it would need to be at least quintic.
If you move the centre to 0,0, i.e. f(x)=1/2+g(x-1/2), then g(x) is odd and so consists only of odd-degree terms:
g(x) = ax^(5) + bx^(3) + cx
f(0)=0 => g(-1/2)=-1/2
f(1)=1 => g(1/2)=1/2
f(1/2)=1/2 => g(0)=0 (this is guaranteed for an odd function)
f'(0)=f'(1)=0 => g'(-1/2)=g'(1/2)=0 (g' is even so g'(-1/2)=g'(1/2) is guaranteed).
g'(x) = 5ax^(4) + 3bx^(2) + c
g'(1/2) = 0 => 5a+12b+16c = 0
f'(1/2) = α => g'(0)=α => c=α (I've used α here to avoid the conflict with the coefficient a).
=> 5a+12b+16α = 0 => a = -(16α+12b)/5
(If you think what happens to the curve outside [0,1], you'll probably want a to be negative).
=> g(x) = -((16α+12b)/5)x^(5) + bx^(3) + αx
g'(x)=0 => x^(2) = 1/4 or x^(2) = -α/(4α+3b)
I'm assuming that you want the function to be monotonic increasing, in which case -α/(4α+3b) should be negative (so that g'(x)=0 only has two real roots) => 4α+3b>0 => 3b>-4α => b>-(4/3)α.
Aside from that, try different values of b to get the desired shape.