Let $T:[0,1]\to[0,1]$ be a monotone and continuous function, and suppose that $\forall x\in[0,1]:T(x)=\frac{f(x)+1-f(1-x)}{2}$.
- Is there a unique monotone and continuous function $f$ that satisfies the above?
- Can we express $f(x)$ explicitly (as a function of $T$)?
For example, if $T(x)=ax+\frac{1-a}{2}$ for some $a>0$, then $f(x)=T(x)-T(0)=ax$ works.
As @Clement answered, if $f$ satisfies the condition, then so will $f+c$ for some constant $c$.
Is $f$ unique up to an additive constant?
This is definitely not unique, as for any $a \in \mathbb{R}$, we have that if $f(x) = x + a$ then $T(x) = x$.
It is not unique up to additive constant either. The idea is that if $f(x) = f(1 - x)$, then $T(x) = \frac{1}{2}$ necessarily. For instance, consider: $$ f(x) = ax^n + a(1 - x)^n $$ where $a \in \mathbb{R}$ and $n \in \mathbb{N}$. Then for any of such $a,n$, we have that $T(x) = \frac{1}{2}$.
EDIT: There also exists $T$ such that no such $f$ exists. One example is the case where $T(x) = 0$. Then this reduces to the problem of solving for: $$ f(1 - x) = f(x) + 1 $$ But letting $x = 0$ gives $f(1) = f(0) + 1$, and letting $x = 1$ gives $f(1) = f(0) - 1$, a contradiction. This probably suggests that $f$ cannot be expressed in terms of $T$.