$f(x) + f(1-x) = f(1)$

Let $g(x) = f(x + 1/2)$, which is defined on $[-1/2, 1/2]$ and which satisfies $g(x) + g(-x) = g(1/2)$. Now, any function $g$ on $[-1/2, 1/2]$ can be written

$$g(x) = \frac{g(x) + g(-x)}{2} + \frac{g(x) - g(-x)}{2} = \frac{g(1/2)}{2} + \frac{g(x) - g(-x)}{2}$$

where the first term is the even part and the second term is the odd part, both of which can be chosen arbitrarily. The problem conditions stipulate the even part and the odd part $h(x)$ is subject to the condition $h(1/2) = \frac{g(1/2)}{2}$. So we can pick an arbitrary odd function for $h$ and then everything else is determined, e.g. if $h(x) = x^3$ then $g(x) = \frac{1}{8} + x^3$.

So the meta-problem is "what conditions on an odd function are strong enough to make it equal to $h(x) = x$?" and I don't think this is a reasonable or interesting question to ask unless you have something very specific in mind.

This problem is not actually much like the Cauchy functional equation because there is only one free parameter instead of two.

I presume $\displaystyle f(1) \neq 0$, in which case we can assume $\displaystyle f(1) = 1$.

If $\displaystyle f$ is such a function, then $\displaystyle g(x) = \sin^{2}\left(\frac{\pi f(x)}{2}\right)$ is also such a function.

If $\displaystyle f(1) = 0$, take $\displaystyle g(x) = \sin(f(x))$.

So you will really need much stronger restrictions than infinite differentiability etc.

As to your edit, continuity is enough.

We can first show $\displaystyle f(1/q) = f(1)/q$ for integral $\displaystyle q$.

This can easily be extended to $\displaystyle f(p/q)$ (for instance $\displaystyle f(2/q) + f(1/q) + \dots + f(1/q) = f(1)$) and by continuity, to the whole of $\displaystyle [0,1]$.