$$(1) F:[0,1] \rightarrow\, [0,1]\,$$
Where a $F$ is a $(C)$ continuous convex function $$\forall t\in [0,1];\forall(x,y)\in[0,1];\,F(tx+(1-t)y)\leq tF(x)+(1-t)\times F(y)$$. where $F$ satisfies $(E)$ . $$(E)\,F(1)\geq 1\,\,, F(0)\leq 0\,\,,\&\, \exists \,\,\text{some third interior fixed point} \,m\in\text{dom}(F);\,F(m)\geq m\,\,\text{where}; 0 \neq m \neq 1$$.
(2) $$F:[0,1] \leftrightarrow\, [0,1]\,$$. Where $F$ is a bi-jective self map ,of the unit interval that satisfies. $(A), (B) \land (SC)$ below.
$$(A) \text{F is 1-auto-diffeo-morphophism of unit interval}$$, Once continuous derivatives of $F$ and its inverse function. $$(A)F(0)=0,\, F(1)=1\land F(m)=m\,\text{where} \,m\in (0,1)$$.
$$(SC)\text{and star convex at 0}: $$. $$(SC)\forall(x\in(0,1)):F(tx)\leq t\times F(x)$$.
Can function (1) $F(x)$ meet with a line segment in three distinct points without $F(x)$ being $F(x)=x$?
Answer no. What about star convex function $(2)$ My answer: $$\forall x\in [0,1];x\geq m):F(x)=x$$
Caveat: If ,yes, Suppose we replace $(SC)$ with midpoint -star convexity at $0$
$$\forall(x)\in [0,1]:F(\frac{1}{2}\times x)\leq \frac{1}{2}\times F(x)$$. where $$m\in \text{Q}\cap [0,1]$$?.
I presume, but do know that the answer is also no here as it is for a strictly monotonic increasing that is continousthat is 2- midpoint convex function with three fixed points.
The only possibility is $F(x) = x$. Here is an argument without assuming differentiability. For greater generality, replace $x^* \in (0,1)$ as the third point where $F(x^*) = x^*$ instead of $x^*=1/2$.
By convexity, for any $x$ in $(0,1)$ with $x\ne 1/2$, the incremental ratio $$R(x) = \frac{F(x) - F(1/2)}{x-(1/2)}$$ is increasing, so $R(0) \le R(x) \le R(1)$.
When $F(x) = x$ for $x=0, 1/2, 1$, we have $$R(0) = R(1) = 1$$ which squeezes $R(x)=1$ and implies $F(x) = x$.