I would to solve the following equation either symbolically or numerically, but I am unsure how:
$$x^{ax} = (1-a+x)^{1-a+x} \quad\text{ where }a,x\in (0,1)$$
I have fixed various values for a and plotted this equation. I believe that each a is mapped to a unique x and that x is increasing in a. However, I am unsure how to show any properties about this solution. Any help is appreciated.
You cannot expect a closed form solution for the zero of function $$f(x,a)=(1-a+x)^{1-a+x}-x^{a x}$$ even using special functions.
If you draw the zero level of the contour plot of the function (have a look here), you notice a nice smooth function.
Solving for a few values of $a$, you can generate the following table $$\left( \begin{array}{cc} a & x(a) \\ 0.0 & 0.000000 \\ 0.1 & 0.079640 \\ 0.2 & 0.142752 \\ 0.3 & 0.198384 \\ 0.4 & 0.249508 \\ 0.5 & 0.297633 \\ 0.6 & 0.343677 \\ 0.7 & 0.388258 \\ 0.8 & 0.431833 \\ 0.9 & 0.474761 \\ 1.0 & 0.500000 \end{array} \right)$$
which is "almost" $$x(a)=\frac 12 a^{\frac 34}$$ obtained using a quick and dirty nonlinear regression $(R^2=0.9996)$.
Using this result as a starting point, Newton method should converge quite fast. Let us try for $a=0.345$; the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.225079 \\ 1 & 0.221839 \\ 2 & 0.221838 \end{array} \right)$$
As @abiessu commented, we could also start with $x(a)=\frac a2$ which for the working case would lead to the same number of iterations.
Edit
Since the proposed estimate $x_0=\frac 12 a^{\frac 34}$ looks to be quite good, we could afford one iteration of Newton method and obtain $$x_1=x_0+\frac{(1-a+x_0) \log (1-a+x_0)-a x_0 \log (x_0)}{a \log (x_0)-\log (1-a+x_0)+a-1}$$
$$\left( \begin{array}{cccc} a & x_0=\frac 12 a^{\frac 34} & x_1 & \text{solution} \\ 0.00 & 0.000000 & 0.000000 & 0.000000 \\ 0.05 & 0.052869 & 0.043191 & 0.043191 \\ 0.10 & 0.088914 & 0.079635 & 0.079640 \\ 0.15 & 0.120514 & 0.112422 & 0.112429 \\ 0.20 & 0.149535 & 0.142745 & 0.142752 \\ 0.25 & 0.176777 & 0.171260 & 0.171265 \\ 0.30 & 0.202680 & 0.198381 & 0.198384 \\ 0.35 & 0.227521 & 0.224394 & 0.224396 \\ 0.40 & 0.251487 & 0.249507 & 0.249508 \\ 0.45 & 0.274713 & 0.273879 & 0.273879 \\ 0.50 & 0.297302 & 0.297633 & 0.297633 \\ 0.55 & 0.319332 & 0.320871 & 0.320872 \\ 0.60 & 0.340866 & 0.343674 & 0.343677 \\ 0.65 & 0.361955 & 0.366113 & 0.366118 \\ 0.70 & 0.382643 & 0.388248 & 0.388258 \\ 0.75 & 0.402964 & 0.410130 & 0.410147 \\ 0.80 & 0.422949 & 0.431806 & 0.431833 \\ 0.85 & 0.442623 & 0.453317 & 0.453358 \\ 0.90 & 0.462011 & 0.474700 & 0.474761 \\ 0.95 & 0.481130 & 0.495990 & 0.496078 \\ 1.00 & 0.500000 & 0.500000 & 0.500000 \end{array} \right)$$