1. $ n $ is even. then Betty will win moving symmetric with respect the moves of Aaron. 2. $ n $ is odd. then Aaron will win: initially he'll put a symbol in the first cell from left to right. then using the case $ n $ even Aaron will move symmetric with respect the moves of Betty. if in some moment Betty move in the last cell then... a mistake...... I'll post it later with the strategy in this case.

can somebody post a solution or at least one hint for the case $ n $ odd?

When n is odd Betty wins.Now,in his first move,Betry places O at the lefrmost or at rhight most cell(at least one is empty).Now,assume that Betty placed O at the leftmost cell(the same with the rightmost).Now,if Aaraon plays at the rightmost at his first or second move,then Betty places O at the right adjacent cell to the one that isn't rightmost,and for every new X he plays O at an adjacent right cell.Now,if A doesn't play at the rightmost in his 2 first moves,Betty at his second move plays the strategy above,expect if player Aaron plays at the rightmost,then he plays at the adjacent one to the cell where Aaraon played first and then continues the sttategy.It is obvios that this strategy works for all n,so no need for an even and odd case