Step 1: Trace where each digit needs to go.
Start with 6 5 3 1 2 4 at positions 1 to 6, target 1 2 3 4 5 6. Position 1 holds 6, which belongs at position 6. Position 6 holds 4, which belongs at position 4. Position 4 holds 1, which belongs at position 1. This closes a loop through positions 1, 6 and 4. Separately, position 2 holds 5, which belongs at position 5, and position 5 holds 2, which belongs at position 2, closing a second loop through positions 2 and 5. Position 3 already holds the correct digit, 3, and forms a loop of its own.
Step 2: Turn the loops into a step count.
A closed loop covering $k$ positions can always be sorted in $k - 1$ swaps when any two positions can be swapped directly, and never in fewer. The 3 position loop (1, 6, 4) needs 2 swaps, the 2 position loop (2, 5) needs 1 swap, and the 1 position loop (3) needs 0 swaps. Total minimum steps:
\[ 2 + 1 + 0 = 3 \]
Step 3: Confirm with an actual sequence of swaps.
Swap positions 1 and 4 (values 6 and 1): 6 5 3 1 2 4 becomes 1 5 3 6 2 4.
Swap positions 2 and 5 (values 5 and 2): 1 5 3 6 2 4 becomes 1 2 3 6 5 4.
Swap positions 4 and 6 (values 6 and 4): 1 2 3 6 5 4 becomes 1 2 3 4 5 6.
The target is reached in exactly 3 steps.
Final Answer:
The minimum number of random switch steps needed is 3.
\[ \boxed{3} \]