Compound with molecular formula C$$_3$$H$$_6$$O can show:

The molecular formula C$$_3$$H$$_6$$O has a degree of unsaturation (index of hydrogen deficiency) of $$\dfrac{2(3) + 2 - 6}{2} = 1$$. This means one degree of unsaturation, which could be a double bond or a ring.

The possible structural isomers include: propanal (CH$$_3$$CH$$_2$$CHO, an aldehyde), propan-2-one or acetone ((CH$$_3$$)$$_2$$CO, a ketone), and allyl alcohol (CH$$_2$$=CHCH$$_2$$OH, an unsaturated alcohol), among others like oxetane (a cyclic ether) or cyclopropanol.

Propanal and acetone are related by functional group isomerism (aldehyde vs. ketone). Allyl alcohol is also a functional group isomer of both (unsaturated alcohol vs. carbonyl compound). Since the compound can exist as both an aldehyde and a ketone (different functional groups with the same molecular formula), C$$_3$$H$$_6$$O exhibits functional group isomerism.

Positional isomerism requires at least two positions for the same functional group (e.g., propan-1-ol and propan-2-ol), but here with only 3 carbons, the aldehyde can only be at the terminal position, and there is only one ketone possible. Metamerism requires different alkyl groups on either side of a functional group like an ether, but with C$$_3$$H$$_6$$O the only acyclic ether would be methyl vinyl ether (only one possibility).

The answer is $$\boxed{\text{Functional group isomerism}}$$.