a) Let $A$ be the set of integers where this trick is possible. We have $2 \in A$, for example by the following strategy: if the first coin is heads, this means the number is $1$, otherwise it is $2$; it is clear that the assistant can choose the status of first coin as he wishes by either turning that coin or the second one. We now show the stronger property $m,n \in A \implies mn \in A$: A coin can be seen as an element of $\mathbb Z/2$, $0$ meaning heads and $1$ meaning tails. Then any row of $n$ coins corresponds to an element of $(\mathbb Z/2)^n$. We also use the shortcut $[N]$ for the set $\{1,2,...,N\}$. Assume the assistant is given $m \cdot n $ coins, then instead of them being arranged in a row, he can assume they are arranged in a rectangular shape of sides $m$ and $n$, i.e. the given constellation is some $(a_{i,j})_{i \in [m],j\in [n]}$ with $a_{i,j}\in \mathbb Z/2$. Let $b_i := \sum_j a_{i,j}$ and $c_j:= \sum_i a_{i,j}$. Then $(b_i)_{i\in [m]}$ and $(c_j)_{j \in [n]}$ are each an arrangement of $m$ and $n$ coins, respectively. By assumption, the magician and the assistant now can agree on some strategy that encode a given $x \in [m]$ in the coins $(b_i)$ by swapping exactly one of them, say the one with index $I$. Similiarly, he can encode an $y \in [n]$ by swapping a coin from the $(c_j)$, say the one with index $J$. Now the assistant simply swaps the coin $a_{I,J}$, which induces exactly the desired swap in the $(b_i)$ and $(c_j)$; thus the magician can revert this procedure and reconstruct the pair $(x,y) \in [m] \times [n]$. In other words, the assistant can always send him a given pair $(x,y) \in [m] \times [n]$; but fixing any bijection between $ [m] \times [n]$ and $[mn]$, this shows that the assistant can send him a number from $[mn]$ as well. b) We claim that this is only possible for powers of of $2$ (the sufficiency of this follows inductively from a)). We say that $x,y \in (\mathbb Z/2)^n$ are neighbours and write $x \sim y$ if they differe at exactly one position; in terms if rows of coins this means they differe by exactly one swap of a single coin. Now every $x$ has exactly $n$ neighbours. For the magician and the assistant to be able to do the trick, they need to define a function $f: (\mathbb Z/2)^n \to [n]$ (to be understood as: if the magician comes in and sees arrangement $x$, he proclaims that $f(x)$ is the number to be encoded) such that: for any given $x \in (\mathbb Z/2)^n$ and any $m \in [n]$ there is a $y \sim x$ with $f(y)=m$. But by every $x$ only having $n$ neighbours and there always being $n$ possible $m$ to be encoded, every $x$ must have exactly one neighbour for each $m$. Thus the following construction is well-defined: For any $x \in (\mathbb Z/2)^n$, there is exactly one neighbour $y$ such that $f(y) \equiv f(x)+1 \mod n$; we call $y=s(x)$ the successor of $x$. Note that $s$ is bijective. Now starting at any $x_0 \in (\mathbb Z/2)^n$, we get a sequence $x_i$ by $x_{i+1} := s(x_i)$. This sequence must eventually get periodic, and by bijectivity will even reach $x_0$ again. Thus there is some period $p>0$ with $x_p=x_0$. But now a trivial induction argument gives $f(x_i) \equiv i + f(x_0) \mod n$, implying $p \equiv f(x_p)-f(x_0) = 0 \mod n$, thus $p$ is divisible by $n$. Now these "cycles" divide $(\mathbb Z/2)^n$ into equivalence classes, each the size a multiple of $n$, thus implying $n|2^n$ being a power of $2$.