Chemistry - Which of the compounds give the same SN1 and SN2 products?

For starters, none of your products show chirality in any way, so it does not matter whether stereoinformation is lost or whether you convert a racemic mixture into a racemic mixture where every single molecule if you had labelled it pre-reaction now had the opposite stereochemistry — in both cases, both reactants and products are racemic mixtures. So you cannot take the presence or absence of chiraliy to distinguish here.

What we can say, is that every compound can take part in an $\mathrm{S_N2}$ reaction (some faster, some slower) but not every one will easily take part in an $\mathrm{S_N1}$ reaction. Specificly, (a) and (d) would give a secondary carbocation intermediate while (b) would even give a primary carbocation. These cations are very unstable and would immediately undergo Wagner-Meerwein type rearrangements towards more stable — read tertiary or resonance stabilised — carbocations. Or, to quote one of my professors from uni:

Exactly one single unstabilised secondary carbocation exists, and that is the prop-2-ylium cation. (Because it cannot rearrange to a more stable one.)

Thus, only (c) will form a stable carbocation without any rearrangement happening: It would be an allyl cation which is well resonance stabilised. But now we have to dig a step deeper, because while all other reactants will certainly not give the same product in both cases, it could be that the correct answer is ‘none of these’, i.e. that somehow we can distinguish between products — because once we have the allyl cation, we could let the nucleophile attack on carbon 1 or carbon 3. Thankfully, though, both attacks give the same product here, because there are no other asymmetries present in the molecule, so we can transform one into the other by rotation.

And, never forget, the $\mathrm{S_N2}$ reaction could also follow the $\mathrm{S_N2'}$ mechanism — this, too, would switch the identities of the two carbons, so we can truly say that both mechanisms will lead to the same product.