instant current in very long transmission line
From your description, yes, initially you see the line-charging current of 5v/50_ohm or 0.1 amp current.

After the initial edge has propagated the entire length of the line and arrived at the (non-matched) 1,000 ohm lumped resistor load, the current into the resistor is only 5volt/1,000_ohm or 5milliamp. There will be a reflection, because the extra current, the 0.095amp, is still flowing; the differential-equation solution has boundary conditions on the energy and these boundary conditions require a large ( 95% ) overshoot on the voltage. However the reflected wave imposes a total current of 5v/1,000_ohm or 0.005 amp, and as that reflected wave propagates back toward your current-meter you STILL read 0.1 amp up until the reflected wave arrives back at the source.

Then your current meter changes readings, from 0.1 amp to 0.005 amp; that change occurs after a total delay of TWICE the electrical length of your cable.

Notice we've not discussed what happens when the reflected wave has arrived back at your 5 volt source. If that source has 50 ohm Rout, what happens?
If that source has 0 ohm Rout, what happens?

===========================

for other description, and perhaps better description than I've written, search for 'Bergeron model'; this model was invented by a French hydraulic engineer, seeking to understand water-hammer (water being a non-compressible low-loss medium of energy transmission).