If a closed smooth manifold $M$ admits a smooth free involution $T$, then it bounds. In fact, the mapping cylinder of the quotient map $M \to M/T$ is the manifold whose boundary is $M$.
Is the converse true? If not, then could someone give an example of a closed smooth manifold which bounds but does not admit any free involution.



Does a manifold which bounds always admit a free involution?
The answer is no. In your example, $M$ is the boundary of a twisted $I$-bundle over another manifold. That's the level of generality in which there's an if and only if statement. 


If you want a closed smooth manifold which bounds, but which does not admit a free involution (you can go further and say it does not admit *any* involution, regardless of having a fixed point set or not) then take a compact orientable hyperbolic 3-manifold that has no symmetries. These exist, since Sadayoshi Kojima proved there is a hyperbolic 3-manifold whose isometry group (therefore the group of homotopy-classes of homotopy equivalences by Mostow rigidity) is any finite group. But all compact 3-manifolds are the boundary of some 4-manifold, this is the Dehn-Lickorish theorem. 


