In organic chemistry, compounds having the same molecular formula but different number of carbon atoms (alkyl groups, etc.) on either side of functional group (i.e., $\ce{-O-,-S-, -NH-, -C(=O) -,}$ etc.) are called metamers and the phenomenon is called metamerism (Wikipedia). Thus, you agree that all three compounds are metamers.
Positional isomers are constitutional isomers that have the same carbon skeleton and the same functional groups (and/or substituents). In Positional Isomerism, the prime carbon chain remains the same in all aspects while the position of functional groups on and/or other substituents change position on a parent structure.
In chain isomerism or skeletal isomerism, components of the (usually carbon) skeleton are distinctly re-ordered to create different structures (Wikipedia).
Accordingly, A and B show metamerism and positional isomerism (position of $\ce{C=O}$ group change on $\ce{C-C-C-C-C}$ carbon skeleton).
A and C show metamerism, chain isomerism ($\ce{C-C-C-C-C}$ carbon skeleton changes significantly to $\ce{C-C-C(-C)-C}$ skeleton), and positional isomerism (position of $\ce{CH3}$ group changes on $\ce{C-C(=O)-C-C}$ carbon skeleton from 1-to-3 position).
Similarly, B and C also show metamerism, chain isomerism ($\ce{C-C(=O)-C-C-C}$ carbon skeleton changes significantly to $\ce{C-C(=O)-C(-C)-C}$ skeleton), and positional isomerism (position of $\ce{CH3}$ group changes on $\ce{C-C(=O)-C-C}$ carbon skeleton from 4-to-3 position).
Therefore, I also agree with you that B and C should show chain isomerism. If not, then n-pentane and 2-methylbutane would not have chain isomerism! :-)
Late Addition: Example for positional isomerism with substitution:
