Assuming that $f(x) > x$ rather than $f(x) \geq x$ for all $x$, and also that $f$ is strictly increasing, one can obtain a $g$ which piecewise has roughly similar properties to $f$, e.g. if $f$ is smooth and has no critical points then $g$ is piecewise smooth with no critical points (something funny happens near critical points though that I don't fully understand).

Firstly, define $x_n$ iteratively for $n=0,1,\dots$ by $x_0 := 0$, $x_{n+1} := f(x_n)$, then the $x_n$ will be increasing to infinity (they cannot accumulate at any finite point $x_*$ as one would then have $f(x_*)=x_*$).  

Now also define $x_n$ for half-integers $n=1/2,3/2,\dots$ by picking $x_{1/2}$ arbitrarily between $x_0,x_1$ and then setting $x_{n+1} := f(x_n)$ for $n=1/2,3/2,\dots$.  The $x_n$ for half-integer $n$ interlace between the $x_n$ for integer $n$, so the $x_n$ still increase to infinity as $n$ ranges over the combined index set $0,1/2,1,3/2,\dots$.

For any $n$ in this combined index set, the function $f$ is a continuous increasing map from $[x_n,x_{n+1/2}]$ to $[x_{n+1},x_{n+3/2}]$ that maps endpoints to endpoints, and must therefore be a bijection (by the intermediate value theorem) and thus a homeomorphism (as both domain and range are compact Hausdorff).  Call $f_n: [x_n,x_{n+1/2}] \to [x_{n+1}, x_{n+3/2}]$ the restriction of $f$ to these intervals.  To finish the job it will suffice to find increasing homeomorphisms $g_n:[x_n,x_{n+1/2}] \to [x_{n+1/2},x_{n+1}]$ mapping endpoints to endpoints such that $f_n = g_n \circ g_{n+1}$ for all $n=0,1/2,\dots$.  It is easy to describe the general solution to this: pick an increasing homeomorphism $g_0: [x_0,x_{1/2}] \to [x_{1/2},x_1]$ mapping endpoints to endpoints, and set $g_n := f^n \circ g_0 \circ f^{-n}$ for integer $n$ and $g_n := f^{n+1/2} \circ g_0^{-1} \circ f^{1/2-n}$ for half-integer $n$.

The function $g$ produced here by gluing together the $g_n$ will be continuous and strictly increasing.  If $f$ has no critical points then the $f_n$ will be diffeomorphisms and thence the $g_n$ will also.  So $g$ will be smooth except at the transition points $x_n$.  One can probably fix things up at the endpoints by imposing appropriate boundary conditions on $g_0$ at $x_0$ and $x_{1/2}$ but I haven't worked these out carefully.  The situation becomes more interestingly complicated when $f$ has critical points $f'(x)=0$ or fixed points $f(x)=x$, but again I haven't looked into this carefully (presumably the analysis of the analytic case in other answers will indicate what the behaviour should be there).