Alice and Bob's spaceships are converging inertially at 0.99c. Alice knows enough about Bob's clock that she can calculate for any time of her clock what his clock would show "at that time" in her frame. Bob is similarly knowledgeable about Alice's clock. When Alice's clock shows Ta, Alice calculates Tb = what Bob's clock shows at that time in Alice's frame. When Bob's clock shows Tb, Bob calculates Tc = what Alice's clock shows at that time in Bob's frame. Question: does Ta = Tc ?
Answer
Short answer: No. If they did, there would be no relativity of simultaneity.
The simplest way to track the conflict is to add a buoy, say in Alice's frame, at some distance from Alice's own location. Let the buoy carry its own clock, synchronized to that of Alice, such that when Bob passes the buoy location, the buoy clock shows time $T_A$, just like Alice's clock. This means Alice observes 2 simultaneous events at non-coincident locations: her clock beating time $T_A$ at her location, and the buoy clock beating time $T_A$ at its own location.
Now look at Bob as he passes the buoy at his time $T_B$ and buoy time (Alice's time) $T_A$. Relativity of simultaneity means that Bob cannot see both Alice and the buoy marking time $T_A$ at the same moment of his time: spatially separated events that look simultaneous to Alice never appear simultaneous to Bob. So although he does see the buoy clock showing $T_A$ at his time $T_B$, he finds that Alice's clock shows a different time ${\tilde T}_A$.
Fun exercise: Apply the same reasoning to a symmetrical setup using buoys in both Alice's and Bob's frame.
Note following comments: As @dmckee observed, the phenomenon is independent of a particular choice of time origin, in either frame. Alice and Bob may choose their individual $t=0$ marks arbitrarily, yet they always arrive at the same conclusion regarding the specific time intervals. In more formal if obvious terms, time intervals are invariant under time translations in any frame.
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