Retract embedding of $S^{n}$ in its unit tangent bundle Acording to the comment of Mark Grant and the answer of Ryan Budney, I revise the question:
For what even $n$, there is  a  retract embedding of of $S^n$ in its unit tangent bundle? 
 A: Sure.  $S^2$ embeds in its unit tangent bundle.  
The unit tangent bundle of $S^2$ is the 3-manifold $SO_3$.  $S^2$ is the boundary of any smoothly embedded $D^3$.  It can't be a retract since $\pi_2 SO_3$ is trivial. 
I think this argument holds in every even dimension, except the relevant homotopy group might be torsion. Think of the unit tangent bundle as a Stiefel manifold of 2-frames, and look up the computation of the first non-trivial homotopy group.  It follows from that.  The only geometric information you need is the clutching map for the tangent bundle of a sphere. 
A: Here is a homological argument. The unit tangent bundle is a spherical fibration 
$$
S^{n-1} \to E \stackrel{p}{\to} S^n
$$
and so it has a long exact Gysin sequence in homology, part of which looks like
$$
\cdots \to H_1(S^n) \to H_n(E)\stackrel{p_\ast}{\to} H_n(S^n) \stackrel{\cap e}{\to} H_0(S^n)\to \cdots .
$$
If $n$ is even, the Euler class $e\in H^n(S^n)$ is twice a generator, and so the final map in the sequence can (after choosing suitable generators) be thought of as multiplication by 2, therefore injective. It follows that $H_n(E)$ is zero. This rules out having a retraction $r: E\to S^n$, which would have to be onto in homology.
