Space of embeddings of circle in a surface Let $S$ be a compact oriented surface of genus at least $2$ (possibly with boundary).  Let $X$ be a connected component of the space of embeddings of $S^1$ into $S$.
Question : what is the fundamental group of $X$?  My guess is that the answer is $\mathbb{Z}$ with generator the loop of embeddings obtained by precomposing the base embedding with a sequence of rotations of $S^1$.
I'm also interested in the higher homotopy groups of $X$, which I would guess are trivial.

Edit: In response to Sam Nead's question, I'm most interested in the smooth category, but am also interested in the topological category.  There are technical issues in giving an appropriate topology to mapping spaces in the PL category, so the question doesn't really make sense there.
 A: For $S$ the sphere, assuming smooth embeddings, any curve divides the sphere into two discs, hence is diffeomorphic to the equator. Then $X$ is a quotient space of the orientation-preserving diffeomorphism group of the sphere by the subgroup that preserves the equator. The orientation-preserving diffeomorphism group of the sphere is homotopic to $SO_3$. The subgroup that preserves the equator is a product of two copies of the group of diffeomorphisms of the disc that fix the boundary. This is known to be contractible.
Thus the space is diffeomorphic to $SO_3$, which is the unit tangent bundle on $S^2$.
So as Sam Nead suspected, there is a lot of higher homotopy.
A: Edit - 
As pointed out by Igor below, my "proof" of the first bullet point is incomplete.  I'll leave the rest of the post here: perhaps some kind soul will fix the gap. 
Original - 
Let's restrict attention to the case where all embeddings are smooth and where $S$ has negative Euler characteristic, or is an annulus or Möbius band. 


*

*If $X$ contains a trivial curve, then $X$ is homotopy equivalent to the unit tangent bundle to the surface $S$. 

*If $X$ contains an essential curve, then $X$ is homotopy equivalent to a circle.


Both of these results follow from Matt Grayson's curve-shortening flow (Annals, 1989).  If $S$ is the torus or Klein bottle, then first bullet still holds, but the second does not.  Instead $X$ is homotopy equivalent to (a cover of) $S$.  Finally, if $S$ is the sphere or projective plane, then I believe that $X$ has nontrivial higher homotopy groups. 
If the embeddings are only continuous, well, that seems tricky. 
A: $
\newcommand{\Homeo}{\operatorname{Homeo}}
\newcommand{\SO}{\operatorname{SO}}
$Since you are interested in the topological category, then I think it will suffice to prove the necessary facts about $\Homeo_0(S)$ and about the curve stabilizer.  Now, there is a topological proof that $\Homeo_0(S)$ is contractible -- see this mathoverflow question: 
Homotopy type of set of self homotopy-equivalences of a surface
Here is the paper of Hamstrom that they are referring to. 
http://0-projecteuclid.org.pugwash.lib.warwick.ac.uk/euclid.ijm/1256054895
Reading a bit of that I learned that Kneser was the first to prove (in 1926!) that $\Homeo_0(S^2)$ deformation retracts to $\SO(3)$.
