$$\alpha(s)=\bar{\alpha}(h(s))$$
Differentiate w.r.t. $s$
$$\alpha'(s)=\bar{\alpha}'(h(s))h'(s)$$
Take absolute value
$$|\alpha'(s)|=|\bar{\alpha}'(h)||h'(s)|$$
$|h'(s)|=1$
Here $h(s)$ should be continuous, otherwise $\alpha$ won't satisfy as a regular curve
$h'(s)=1$ or $-1$
$h(s)=\pm s +s_0$ for some constant $s_0$
Arclength parametrization only means $|\alpha'(s)|=1$ at each point on the curve, it doesn't tell us anyone about the orientation. Reversing the direction would still give us the arclength parametrization.
If you prolong the axes of the rear and of the front wheels, the point where they meet is the center of rotation.
Since the rear axle is fixed and steering is obtained by rotating the whole front axle, we get a scheme as the one shown.
It is clear from the picture that, when the steering angle $\beta$ has a large value, the rear and front turning radii
are sensibly different: an effect that is well learnt by fresh drivers when parking backwards.
The (ideal) kynematics of the car obeys to the following set of equations
$$ \bbox[lightyellow] {
\left\{ \matrix{
{\bf u}_{\,T} = \left( {\cos \alpha ,\sin \alpha } \right)\quad {\bf u}_{\,H}
= \left( {\cos \left( {\alpha + \beta } \right),\sin \left( {\alpha + \beta } \right)} \right) \hfill \cr
{\bf n}_{\,T} = \left( { - \sin \alpha ,\cos \alpha } \right)\quad {\bf n}_{\,H}
= \left( { - \sin \left( {\alpha + \beta } \right),\cos \left( {\alpha + \beta } \right)} \right) \hfill \cr
{\bf H} = {\bf T} + L\,{\bf u}_{\,T} \hfill \cr
0 = {\bf n}_{\,T} \cdot \,{d \over {dt}}{\bf T} = {\bf n}_{\,H} \cdot {d \over {dt}}{\bf H}
= {\bf n}_{\,H} \cdot \left( {{d \over {dt}}{\bf T} + L\,{d \over {dt}}{\bf u}_{\,T} } \right) \hfill \cr} \right.
}$$
where $\alpha$ and $\beta$ and the other parameters are function of time.
Now, if we assume that the car has a rear traction which provides, through a differential gear, a constant
average speed $v$ at the center of the axle $T$, then
$$
{d \over {dt}}{\bf T} = v\,{\bf u}_{\,T}
$$
and we get an equation linking $\alpha$ and $\beta$ as
$$
\eqalign{
& 0 = {\bf n}_{\,H} \cdot \left( {{d \over {dt}}{\bf T} + L\,{d \over {dt}}{\bf u}_{\,T} } \right) = \cr
& = \left( { - \sin \left( {\alpha + \beta } \right),\cos \left( {\alpha + \beta } \right)} \right) \cdot \left( {v\,\left( {\cos \alpha ,\sin \alpha } \right)
+ L\left( { - \sin \alpha ,\cos \alpha } \right){{d\alpha } \over {dt}}} \right) = \cr
& = - v\sin \left( {\alpha + \beta } \right)\cos \alpha + L{{d\alpha } \over {dt}}\sin \left( {\alpha + \beta } \right)\sin \alpha + \cr
& + v\cos \left( {\alpha + \beta } \right)\sin \alpha + L{{d\alpha } \over {dt}}\cos \left( {\alpha + \beta } \right)\cos \alpha = \cr
& = - v\sin \beta + L{{d\alpha } \over {dt}}\cos \beta \cr}
$$
i.e.:
$$ \bbox[lightyellow] {
{{d\alpha } \over {dt}} = {v \over L}\tan \beta
}$$
For a constant $\beta$ and $\alpha(0)=0$ we get
$$ \bbox[lightyellow] {
\left\{ \matrix{
\alpha (t) = \left( {{v \over L}\tan \beta } \right)t = \omega \,t \hfill \cr
{d \over {dt}}{\bf T} = v\,{\bf u}_{\,T} = v\,\left( {\cos \left( {\omega \,t} \right),\sin \left( {\omega \,t} \right)} \right) \hfill \cr
{\bf T}(t) = {\bf T}_{\,0} + {v \over \omega }\left( {\sin \left( {\omega \,t} \right),1 - \cos \left( {\omega t} \right)} \right) \hfill \cr
{\bf H}(t) = {\bf T}(t) + L\,\left( {\cos \left( {\omega \,t} \right),\sin \left( {\omega \,t} \right)} \right) \hfill \cr} \right.
}$$
The trajectories of $T$ and $H$ are concentric circles
Best Answer
Yes, Franz Wegner constructed pairs of smooth closed curves that are not circles and can serve as pairs of bicycle tracks traversed in either direction. They can be expressed analytically in terms of Weierstrass's $\sigma$ and $\zeta$ functions. Interestingly enough such curves also describes shapes that can float in any position, and trajectories of electrons moving in a parabolic magnetic field.
Short description and a picture are here http://www.tphys.uni-heidelberg.de/~wegner/Fl2mvs/Movies.html#animations, mathematical details and more pictures here http://arxiv.org/pdf/physics/0701241v3.pdf.