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? 



Retract embedding of $S^{n}$ in its unit tangent bundle
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.


