Identical Geodesics implies scalar multiple of metric? Suppose $(M,g^1)$ and $(M,g^2)$ are two intrinsic metric spaces with the same underlying set $M$.  
Assume that for every $p,q\in M$, for each geodesic $\gamma^1_{[p,q]}$ connecting $p$ to $q$ under $g^1$ there exist a geodesic and $\gamma^2_{[p,q]}$ connecting p and q and $g^2$ identical to it, and visa-versa.  
Does this imply that $g^1$ is a multiple of $g^2$?
 A: No. Consider as your model set $M$ a tripod. (A tripod is a graph with one vertex of degree three and three vertices of degree one attached to it.)
For your different length metrics, just assign the edges of the tripods various different lengths, and equip the space with the length metric induced by those edge lengths (making each edge into an isometric copy of $[0,L]$ where $L$ is its length.)
Any two such tripods have the same setwise geodesics, but need not have the same metrics, even up to scaling.
A: Your notation suggested another alternative:  Let $g^2 = (g^1)^2$ -- a local extremum of $g^1$ is also a local extremum of its square.  In fact, let $\phi$ be strictly increasing on $[ 0, \infty )$, then $g^2 = \phi(g^1)$ replicates the extrema of $g^1$.  So geodesics (paths whose length is first order stationary under perturbation, hence correspond to local extrema) are preserved under any such $\phi$.
A: While Anon's answer is correct, here is an example from Riemannian geometry. Start with $(M,g)$ equal to the projective plane with the standard metric of constant curvature, so that geodesics are projective lines. Let $f: M\to M$ be a generic projective transformation and $g'=f^*(g)$ be the pull-back of $g$. Then the two Riemannian manifolds $(M,g), (M,g')$ have exactly the same set of geodesics but are not even conformal to each other.  
