Tiffeneau–Demjanov rearrangement products
Quite a fun one to work out. The first thing to know is that there is a stereoelectronic requirement for the migrating group to be anti to the leaving group ($\ce{N2+}$), for orbital overlap reasons.



(apologies for hand-drawing - doing this on my phone)

You need to consider the chair conformations of both stereoisomers, in both possible ring-flip forms. For the cis stereoisomer, you would expect roughly a 1:1 mixture of ring-flip forms. When the $\ce{OH}$ is axial, the group anti to the LG is H, and the cyclohexanone is formed. Similarly, when $\ce{OH}$ is equatorial, the $\ce{C-C}$ bond is anti and the cyclopentane product is formed.

Now, for the trans-isomer there are two factors favouring the cyclopentane. First, the chair form in which both groups are equatorial will predominate, and this is the stereoisomer that leads to the cyclopentane. Secondly, in the (minor) conformer where the $\ce{OH}$ and $\ce{N2}$ are axial, there is no anti group to migrate according to the normal mechanism. The OH itself is anti to the $\ce{N2}$, and I'm not aware (off the top of my head) of any examples of 1,2-hydroxy shifts. However, I expect some epoxide might be formed via an SN2 pathway.