The key is to figure out the solution for $f(x) = x$ and have the audacity to generalize that. Just as another hint, don't think too hard about polynomials.

Does anyone have a full solution to this problem? Thanks!

First prove that a quadratic $g$ works for $f(x)=x$ and then generalize to any $f$.

Can anyone give a full solution to this Thanks?

Suppose we have such a f. Pick your top 2017 favorite numbers(which is just another way of saying they don't matter) and apply the Lagrange Interpolation Formula and the defining equation to find g. Does this count as a sol?

duck_master wrote: ... This only shows the existence of $m$ but does not find it. We claim that $m = 2$. To see that $m = 1$ will not work, suppose we have some linear $g(x) = ax + b$ satisfying this condition and $f(x) = x$. Then $\max a_i \max f(a_i) = \max g(a_i) = g(\max a_i)$ and similarly for minimum, hence $g(x) = x$, contradiction. Now for any $f$, pick an interval $[M,\infty)$ over which $f$ is strictly increasing (or strictly decreasing, but the two cases can be dealt with the same way). Pick two points $(x_1, f(x_1)), (x_2,f(x_2))$ with $23333M^{23333} < x_1 < x_2$, then choose some concave quadratic $g$ that passes through these two points. For some suitable $g$, we will choose $r \in (x_1,x_2)$ such that $g(r) > f(x_2) > f(x_1) > gf^{-1}g(r)$ (if this cannot be done then decrease the curvature of $g$. Write $g(x) = ax^2 + bx + c$, as $a\to -\infty, g(x) \to a(x-\frac{x_1 + x_2}{2})^2 + f(x_1)$, then choosing $r = \frac{x_1 + 2x_2}{3}$ we see that $g(r) \to \infty, f^{-1}g(r) \to \infty, gf^{-1}g(r) \to -\infty$ as well). After this we construct the following sequence: $r_0 = r,$ and for each $i\ge 1$ let $r_i$ the unique real number within $(x_1, x_2)$ such that $g(r_i) = f(r_{i-1})$. This infinite sequence is decreasing, approaches but always stays above $x_1$. Our problem would be solved if we can find $a_1$ such that $(f^{-1}g)^{(2017)}(a_1) = a_1$ and $(f^{-1}g)(a_1) \ne a_1$. In our case it suffices to find such $a_1$ between $x_1, x_2$ (which are the only two fixed points of $f^{-1}g(x)$). Note that $f^{-1}g$ is continuous over this interval, $(f^{-1}g)^{(2017)}(r_{2016}) = (f^{-1}g)(r_0) > f^{-1}(f(r_0)) = r_0 > r_{2016}$ and $(f^{-1}g)^{(2017)}(r_{2015}) = (f^{-1}g)^2(r_0) < f^{-1}g(x_1) = x_1 < r_{2015}$. Thus there actually exists $x \in (r_{2016},r_{2015}) \subset (x_1,x_2)$ with $f^{-1}g(x) = 0$ by intermediate value theorem, as desired.