= Solution
After compression along $z$, let the original conventional edges be $a,a,c$ with $c=a/\sqrt2$. This gives exactly the specified <volume> reduction. The compressed FCC points admit the alternative translation vectors
$$
\mathbf b_1=\frac a2(1,1,0),\qquad
\mathbf b_2=\frac a2(-1,1,0),\qquad
\mathbf b_3=(0,0,a/\sqrt2).
$$
They are mutually perpendicular, have equal length $b=a/\sqrt2$, and form a right-handed cube. Besides its corners, the lattice has the point
$$
\frac{\mathbf b_1+\mathbf b_2+\mathbf b_3}{2}
=(0,a/2,a/(2\sqrt2)),
$$
which is the compressed image of an original FCC face centre. All compressed FCC sites are generated by the cube's integer translations and this body-centre coset. Therefore
$$
\boxed{\text{compressed FCC}\equiv\text{BCC with conventional edge }b=a/\sqrt2.}
$$
The old compressed tetragonal conventional cell has <volume> $a^3/\sqrt2$ and four sites; the new BCC cube has half that <volume> and two sites. This verifies the same physical site density rather than silently losing atoms during the change of cell. The continuous uniaxial deformation is the <Bain path>, named for <Edgar Bain>.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2014-bain-path-and-cell-reindexing.png]
{title=The <Bain path>: a rotated conventional cell exposes BCC geometry}
{description=Original geometric construction. The basal panel shows the old square and a new square generated by the two diagonal half-translations. The spatial panel uses new-cell coordinates, with corner sites and a body centre. The compressed vertical edge and both new basal edges all have length $a/\sqrt2$. Dots are lattice sites rather than hard-sphere atomic surfaces.}
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