For four frogs, we can win:
1111 -> 0211 -> 0013 -> 0004
11101 -> 02101 -> 03001 -> 00004
11011 -> 02011 -> 02020 -> 04000
111001 -> 201001 -> 003001 -> 000004
1101001 -> 0201001 -> 0003001 -> 0000004
Any other position with four frogs in different places is either a left-right reversal of one of these, or no sequence of moves will result in all frogs in the same place. Hence a(4)=5.
Equivalently, one can start with four frogs in a single place and make a tree of all possible reverse moves, and then count leaves in which all frogs are in different places:
_____________4____
/ | \
___3001___ 202 13
/ | | \ / \ |
201001 12001 2101 1021 1102 112 121
/ | | | |
1101001 111001 11101 1111 11011
Again we see that a(4) = 5. (Note that nodes whose labels or reversed labels have already occurred earlier in a row are omitted from this tree.)