Geodesic representatives in the orbifold fundamental group Does every element in the orbifold fundamental group $\pi_1^{orb}(X,x)$ of a closed hyperbolic 2-orbifold $X$ admit a unique geodesic arc representing it? 
Does every free homotopy class in $X$ admit a unique closed geodesic representing it? 
By a geodesic arc I mean a geodesic loop in the orbifold $X$, whose only possible non-smooth point is the basepoint, whereas a closed geodesic is a geodesic which is smooth everywhere. I know the above facts are true for closed hyperbolic manifolds and I'm wondering about possible analogues in the orbifold setting as I was unable to track down a reference for them.  
 A: First of all, I would use a different name for what you call geodesics on $X$, let's call them regular geodesics. The reason is that geodesics in Riemannian geometry are curves satisfying the equation $\nabla_{c'} c'=0$. For an $n$-dimensional Riemannian orbifold $X$ this equation makes sense not on $X$ itself but in its local coordinate orbifold charts $U$, where 
$$
R^n\supset U \stackrel{\phi}{\to} V\subset X$$
and $\phi$ is the quotient map by a finite group action on $U$. Both Riemannian  metric and (Levi-Cevita) connection $\nabla$ are defined on $U$ and not on $V$. Then a geodesic on $X$ is a curve $\bar{c}$ which admits smooth local lifts $c: I\to U$ satisfying the equation $\nabla_{c'} c'=0$.
For instance, if the Riemannian orbifold $X$ is good and has a Riemannian covering $\tilde{X}\to X$ then geodesics in $X$ are just projections of geodesics in $\tilde{X}$. 
With this in mind, let's say that a geodesic $\bar{c}$ in a Riemannian orbifold $X$ is regular if it is disjoint from the singular locus of $X$. 
Suppose that $X$ is a compact connected hyperbolic orbifold (of any dimension), i.e. it is a Riemannian orbifold of constant curvature $-1$. Then $X$ is complete (as a metric space and also geodesically complete) and good; there are two ways to prove the latter: one uses the fact that $X$ is nonpositively curved and the other uses the fact that $X$ is geometric in Thurston's sense.  The first proof you can find for instance in Ratcliffe's book "Foundations of Hyperbolic Manifolds" and the second proof is due to Matsumoto (early 1990s). 
In any case, the universal cover of $X$ is isometric to the hyperbolic space ${\mathbb H}^n$ and $X$ itself is isometric to the quotient orbifold  ${\mathbb H}^n/\Gamma$ where $\Gamma< Isom({\mathbb H}^n)$ is a discrete cocompact subgroup. A geodesic in $X$ is regular if and only if its lift to ${\mathbb H}^n$ is disjoint from the set of fixed points of elliptic elements of $\Gamma$. 
Every loop $\alpha$ in $X$ is freely homotopic (in the orbifold sense: from now on, homotopy in $X$ is understood only in the orbifold sense) to a (possibly constant!) periodic (closed) geodesic. Similarly, every based loop $\beta$ in $(X,x)$ is homotopic (rel. $x$) to a (possibly constant!) geodesic loop based at $x$. The proofs are straightforward and are the same as in the case of Riemannian manifolds. For based  loops, pick a preimage of $x$, $\tilde{x}\in {\mathbb H}^n$; the relative homotopy class of $\beta$ is represented by an element $\gamma\in \Gamma$, so connect $\tilde{x}$ to $\gamma\tilde{x}$ by a geodesic $c$ in ${\mathbb H}^n$ and project $c$ to $X$. In the case of geodesic loops, you minimize the arc-length among curves in the free homotopy class of $\alpha$. Or use the fact that every element of $\Gamma$ is hyperbolic or elliptic (since $X$ is compact). This proof also shows uniqueness for based loops (because of uniqueness of geodesics in ${\mathbb H}^n$ between the given pair of points). In the case of free homotopy classes you have to be a bit more careful: If the free homotopy class $[\alpha]$ is represented by the conjugacy class of a hyperbolic element $\gamma\in \Gamma$ then uniqueness follows from uniqueness of the geodesic axis $A_\gamma$ of $\gamma$ (the unique $\gamma$-invariant geodesic in ${\mathbb H}^n$). 
In the non-hyperbolic case the situation more subtle. Even if $X$ is a manifold and $[\alpha]$ is trivial free homotopy class then every constant map to $X$ will represent this class! Suppose that $X$ is an orbifold and $[\alpha]$ is represented by a nontrivial elliptic element $\gamma\in \Gamma$. Then every closed geodesic in the class $[\alpha]$ is represented by a constant map to the projection of the fixed-point set of $\gamma$ to $X$. (This projection may consist of more than one point even in the 2-dimensional case, i.e. when $\gamma$ is a reflection.) Of course, such a geodesic will not be regular, hence, you obtain uniqueness of regular closed geodesics in the given free homotopy class. 
Now, to the existence part of your question. As we just saw, if the free homotopy class is elliptic then you cannot have a regular closed geodesic in this class. But even if the free homotopy class is hyperbolic the existence can fail. For instance, suppose that $X$ is the quotient of a hyperbolic surface $Y$ by the hyperelliptic involution $\tau$ ($p: Y\to X=X/\tau$ is the quotient map). The involution $\tau$ anticommutes with some hyperbolic elements $\gamma\in \pi^{orb}_1(X)$, i.e. it sends a homotopically nontrivial loop $\ell$ in $Y$ to itself reversing the orientation. Let $X={\mathbb H}^2/\Gamma$, let $\Gamma_1< \Gamma$ be the index 2 torsion-free subgroup such that $Y\cong  {\mathbb H}^2/\Gamma_1$; let $\sigma\in \Gamma$ be an elliptic element projecting to the involution $\tau$ under the map ${\mathbb H}^2\to Y$. Then $\sigma$ anticommutes with a hyperbolic element $\gamma\in \Gamma$ representing the free homotopy class of $\ell$. Thus, $\sigma$ preserves the geodesic axis $A_\gamma\subset {\mathbb H}^2$ of $\gamma$ and fixes a point $a$ on   $A_\gamma$. Translating this to the geometry of $X$, we see that the unique geodesic $c$ in the free homotopy class of $p(\ell)$ in $X$ passes through a singular point on $X$, i.e. the projection of $a$ to $X$. Therefore, the free homotopy class of $p(\ell)$ contains no regular geodesics. 
