Is there a degree $N$ polynomial $f(x)$ such that
My gut feeling is that there is not, but I don't know how to show that. (Edit: I don't have that gut feeling anymore after being shown it was wrong.)
For $N\geq 3$, there is such a polynomial. Take $f(x)=x^{N-3}(x-3x^2+3x^3)$. To show that this satisfies the desired property, it suffices to check the $N=3$ case.
From here, the $N\geq 3$ case follows since multiplying by the nonnegative increasing polynomial $x$ does not disrupt any of the necessary properties.
For $N=1$, there is no such polynomial. The property that $f(1)=1$ implies that $f(x)$ would have to be $1-a(1-x)$ for some real $a$. If $a>1$ then $f(0)<0$, a contradiction, while if $a\leq 1$ then $f(x)\leq x$ for all $x\in [0,1]$.
For $N=2$, there is no such polynomial. Any quadratic $f$ besides $x^2$ can intersect the graph of $x^2$ at at most two points. One of these points is $x=1$; we know that $f(0)\geq 0$, so the other intersection point $(x_0,x_0^2)$ must satisfy $x_0\in [0,1)$. In particular, $f(x)<x^2$ for $x$ slightly smaller than $1$, meaning that $f'(1)\geq 2$. Also, $f'(0)\geq 0$. Since $f$ is linear, this implies $$f(1)=f(0)+\int_0^1 f(t) dt=f(0)+\int_0^1 \big((1-t)f'(0)+tf'(1)\big)=f(0)+\frac{f'(0)+f'(1)}2.$$ This means $1\geq f(0)+1$, and so $f(0)=0$. Furthermore, as these inequalities are each tight, we must have $f'(0)=0$ and $f'(1)=2$; the only quadratic satisfying these conditions is $f(x)=x^2$.
