Let $n$ be the current number of one of the players. Because both players start with an odd number, $n$ will be odd as it remains positive throughout the game. If $n = \sum_{i=0}^{k} b_i 2^i$ with $b_i \in \{0, 1\}$ is the binary representation of $n$, then $b_0 = 1$. If we assume $b_k = 1$, then the smallest power of $2$ that is bigger than $n$ is $2^{k+1}$, hence, the next number for this player will be
$$2^{k+1} - n = 1 + \sum_{i=0}^{k} 2^i - \sum_{i=0}^{k} b_i 2^i = 1 + \sum_{i=0}^{k} (1 - b_i) 2^i = \sum_{i=0}^{k-1} c_i 2^i,$$
where $c_i = 1 - b_i$ for $1 \le i < k$ and $c_0 = 1$. More visually, the effect of one turn is
$$(1 b_{k-1} b_{k-2} \cdots b_2 b_1 1)^2 \mapsto (\overline{b_{k-1} b_{k-2} \cdots b_2 b_1} 1)^2$$
where $\overline{0} = 1$ and $\overline{1} = 0$. This new number has at most $k$ binary digits, but it may have fewer if $b_{k-1}$ was equal to $1$.
In general, all the leading digits $1$ will be lost, i.e. if $b_k = b_{k-1} = \dots = b_{k-j+1} = 1$ and $b_{k-j} = 0$ then the new number will have $j$ digits less than $n$. Therefore, the number of turns to reach $1$ is equal to the number of blocks of consecutive equal digits in the binary representation of the initial number, where we ignore the last (units) digit in the count.
For example, the smallest number which allows for ten turns is
$$1365 = (10101010101)^2$$
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The smallest number that allows for nine turns is $683 = (1010101011)^2$.
The number $1195 = (10010101011)^2$ allows for nine turns as well.
Abigail can choose $683$ which enables her to carry out nine steps. If Zoe chooses $1195$ she will also be able to carry out nine steps and so she will win.