There is only one way an inertial frame can move relative to another at a given velocity:
(Lemma) If inertial frame K'' is moving at v relative to K, then K' and K'' are at rest relative to each other, and the time values of each pair of co-located clocks of K' and K'' differ by the very same constant.
(Proof) Take a point P in K at time t. The corresponding points P' and P'' of K' and K'', respectively, always coincide in K. So, due to universality (transitivity), the constant velocity at which K' and K'' are moving relative to each other must be 0. This is because at any two t1' and t2' in K', P' corresponds (through transitivity via K) to the same P'' of K'', and similarly, P'' in K'' always corresponds to the same P' of K'. Furthermore, since all clocks are synchronized in both K' and K'', the coordinate times of K' and K'' can differ only by a constant.
Now we can prove an expected result:
(Theorem) v' = v, and thus v' = –v.
(Proof) Let Kw be an inertial frame moving at some speed 0 ≤ w ≤ v relative to K along the x-axis in the positive direction, and let the xw-axis of Kw be defined similarly to that of the x'-axis of K'. Relative to Kw: when w = 0, K' is moving at (positive) speed v, while K at speed 0; when w = v, K' is moving at speed 0, while K at (negative) speed v'. Because of continuity, there exists a w where K and K' are moving relative to Kw at equal speed but in opposing directions. From the symmetry of that situation follows v' = v.
Notes: (a) roughly speaking, "continuity" is the assumption that whenever a parameter is being changed continuously, the result is also changing continuously, (b) the lemma is necessary to establish the symmetry.
In the proof it was tacitly assumed that:
(Assumption) Kw exists for any w < v.
This is not as obvious as it looks; what if v > c, are we sure that w = c would be possible? Nevertheless, we maintain the assumption, as later it will be shown that independently of the above theorem, v < c holds.