What symmetry is responsible for the degeneracy of the free particle Hamiltonian? Consider the hamiltonian $H=\frac{p_x^2}{2m}$ in 1-D. It is invariant under $p_x \rightarrow -p_x$.
Again, this hamiltonian also has translational symmetry. Which one of these two is responsible for doubly degenerate energy eigenfunctions for a given energy $E$? I think it is the first one (Should I call it parity symmetry?). Am I right? 
We know any symmetry appears as a degeneracy in quantum mechanics. Right? Then what is the manifestation of translational symmetry?
 A: You are right concerning the parity transformation, it implies the degeneracy of all states with finite momentum.
The effect of the translation symmetry does not imply more than what is known from the conservation of momentum, as both operators are closely related.
Indeed, the translation operator is given by $$\hat T(a)=e^{i \hat P a}$$ (up to a sign, depending on the active/passive point of view).
One trivially checks that it indeed commutes with the Hamiltonian of a free particle. The eigenstates of the Hamiltonian are the eigenstates of the momentum operator $|p\rangle$, with wavefunction $\Psi_p(x)\propto e^{i p x}$. All translated states $$|p\rangle'=\hat T(a) |p \rangle=e^{ipa}|p \rangle$$ are degenerated. At the level of the wavefunction $$\Psi_p'(x)=e^{ipa}\Psi_p(x)\propto e^{i p (x+a)}$$.
All that only means that the eignestates all have the same energy, independent of the place where one puts the origin.
A: No, the parity symmetry is not the reason free particle states are doubly degenerate, because the group $\mathbb{Z}_2$ is an abelian group, and hence has no irreducible representations of dimension greater than one. The translational symmetry isn't responsible either, for the same reason. You need both.
Suppose that we instead are working with a symmetric potential $U(x) = U(-x)$, so we only have parity symmetry. In this case, the fact that
$$P |p\rangle = |-p\rangle$$
does not produce any degeneracy, since $|p\rangle$ and $|-p\rangle$ are not eigenstates of the Hamiltonian. Instead, what it tells us is that, since we can simultaneously diagonalize $H$ and $P$, we can choose all eigenstates to be even or odd. 
Returning to the free particle situation, this reasoning says that we should be looking at even and odd wavefunctions such as $\cos(px)$ and $\sin(px)$ to find eigenstates, not the plane waves $e^{ipx}$. However, it doesn't tell us that $\cos(px)$ and $\sin(px)$ are degenerate! Instead, their degeneracy follows from the fact that they are linked by translational symmetry,
$$\cos(px) = \sin(px + \pi/2).$$
Doing a change of basis, we recover the fact that $e^{ipx}$ and $e^{-ipx}$ are degenerate.
Mathematically, we can get degeneracy by considering both parity and translational symmetry, because these operations don't commute. The resulting nonabelian group does admit nontrivial irreps, in this case irreps of dimension two. 
For some background on this reasoning, see here. Also note that the argument above doesn't require that the translational symmetry is continuous, so it also applies to (and explains the degeneracy of) Bloch states with opposite crystal momenta.
