Is it true that a map that maps geodesics to geodesics must be an differeomorphism? Let $M,N$ be connected Riemannian manifolds. Let $f:M\rightarrow N$ be a bijective smooth map that maps any unit speed geodesic in $M$ to unit speed geodesic in $N$. 
Question: Is this suffice to show that $f$ induces a differeomorphism $M\cong N$? Can $f$ even be an (local) isometry? If either is not true, what additional assumption is needed (like $M,N$ are simply-connected)? 
(originally asked by a classmate and I was stuck)
 A: Your condition forces $||f_*v|| = ||v||$ for any vector $v$, as pointed out by John in the comments. (Quick proof: given $v$, let $\lambda = 1/||v||$ and consider the geodesic $\exp(t\lambda v)$; $f$ maps this to a unit speed geodesic, so $||f_*(\lambda v)|| = 1$.)
Thus the derivative $f_* : T_pM \to T_{f(p)} N$ is nonsingular at every point $p$, and so the inverse function theorem implies that $f$ is a local diffeomorphism. Since you've assumed $f$ is bijective, $f$ must be a diffeomorphism. $f$ is then evidently also an isometry since inner products are determined by their associated norm.
Supplemental Interesting Stuff About Smoothness:
I believe this is true even if $f$ isn't assumed smooth, that is, a bijective map $f: M \to N$ of connected Riemannian manifolds that takes unit speed geodesics to unit speed geodesics is necessarily a smooth isometry. The rest of this answer will be an outline of a proof of this fact, so if you're not interested you can stop reading now.
Your condition that geodesics go to geodesics certainly forces $d(f(x), f(y)) \leq d(x,y)$, so $f$ is continuous. I claim that $f$ is a local metric isometry in the sense that given $x \in M$, there exists a neighborhood $U$ of $x$ such that for $y, z \in U$ we have $d(f(y), f(z)) = d(y,z)$. 
To see this fix $r$ so that the geodesic ball $B_r(f(x))$ about $f(x)$ of radius $r$ in $N$ is strongly convex in the sense that there is a unique $N$-geodesic between any two of its points living entirely within the ball, and this geodesic is minimizing in $N$. By making $r$ smaller if necessary we may also assume $B_r(x)$ is strongly convex in $M$. Note that $f$ maps $B_r(x)$ into $B_r(f(x))$.
Fix $y,z \in B_r(x)$; let $\gamma$ be the minimizing geodesic from $y$ to $z$. Then $f \circ \gamma$ is a geodesic from $f(y)$ to $f(z)$, which is contained within $B_r(f(x))$, hence is the unique such geodesic, hence is the unique minimizing geodesic between these points. Thus $d(f(y), f(z)) = d(y,z)$ as claimed.
But then fix $s < r$, for $r$ small enough so that the above holds, and consider the restriction of $f$ to the sphere $S_s(x)$ of radius $s$ about $x$. This is homeomorphic to an $(n-1)$-sphere and its image lies in the $(n-1)$-sphere $S_s(f(x))$ about $f(x)$ in $N$. Since $f$ is injective and continuous and $S_s(x)$ is compact, the restriction of $f$ to $S_s(x)$ is a homeomorphism onto its image. But the $(n-1)$-sphere does not embed as a proper subset of itself, so the image of $f$ must contain all of $S_s(f(x))$. In particular, $f$ is a metric-preserving homeomorphism $B_r(x) \to B_r(f(x))$.
Now apply the Meyers-Steenrod theorem to the map $f : B_r(x) \to B_r(f(x))$ to see that $f$ is smooth on $B_r(x)$. 
