Suppose we have a functional equation in the form $$f(1-x)=\chi (x) f(x)$$ with given function $\chi (x)$. What is the condition on the function $\chi (x)$ so that we can write this reflection relation in a symmetric form $$\xi(1-x)=\xi(x) ?$$ I have the only simple answer: the $\chi (x)$ function should be of the type:$$\chi (x)= \phi(1-x)/\phi(x)$$ then $\xi(x)=f(x)/\phi(x)$. Is it correct?
Edit I want to illustrate my question with an example.
Let first $\chi (x)$ be $$ \chi (x)= \frac {\pi}{\sin\pi x (\Gamma (x))^2} $$ And we have functional equation $\Gamma (x) \Gamma (1-x)=\frac {\pi}{\sin\pi x}$ without symmetric form.
Let then $\chi (x)$ be $$ \chi (x)= \frac { \pi^{-\frac {x}{2}}\Gamma(\frac{x}{2})} { \pi^{-\frac {1-x}{2}} \Gamma(\frac{1-x}{2})} $$ Here $\phi(x)= ({ \pi^{-\frac {x}{2}}\Gamma(\frac{x}{2})})^{-1} $ and we have symmetric form with $\xi(x)=\zeta(x)/\phi(x)$
There is not much going on here. Just to make that clear, without loss of essential generality, suppose that we have any function $\ f(x)\ $ and define $\ g(x) := f(x) - f(-x).\ $ Now we have $\ g(-x) = f(-x) - f(-(-x)) = f(-x) - f(x) = -g(x).\ $ Thus, $\ g(x)\ $ is an odd function as in Even and odd functions. In fact, it is twice the odd part of $\ f(x).\ $ Further, it follows that by adding any even function to $\ f(x)\ $ we can recover the same odd function $\ g(x).\ $
In your case, addition and subtraction is replaced by multiplication and division, and also the involution $\ x \to -x\ $ is replaced by $\ x \to 1-x.$