When does hh:mm = mm.ss?

Raku, 14 12 11 bytes


Try it online!

Since we know the bounds of the input, we can substitute a constant operation and a floor on the input in base 60. That number by the way is around 1358/1381, which is the maximum the input differs from the output in base 60. There may be a smaller constant, or at least a smaller way to represent it. For reference, the shortest constant you can multiply by, rather than divide, is 1.01694.

Python 3.8, 40 36 bytes

lambda h,m:(d:=h+(h+m)//60,(d+m)%60)

Try it online!

R, 42 bytes


Try it online!

Version 2: -3 bytes thanks to clarification that we don't need to output the first match when the timer & clock show the same numbers.
So this version outputs the second match for an input of 0:59 (in other words, 1:00instead of the first match at 0:59), and similarly for all other outputs that can end with :59 or :00.

R, Version 1: 49 45 bytes


Try it online!

Outputs the first timer-clock match (so, always a match ending 0:59 instead of a subsequent match ending :00).

This exploits the floating-point rounding of *6/5.9 to slightly less than *60/59, but using the same number of characters. This effectively gives us a floor-like result that rounds-down exact integers in the output (the desired behaviour). Using *60/59 gives an exact floating-point result and so doesn't do this.

(Frustratingly, though, it still isn't as short as simply porting ovs's approach for 43 bytes). Version 2 (above) is shorter.