Unicity of the Perfect Golay Binary Code I am struggling understanding why the Perfect Golay Binary Code is unique (up to equivalence). I found this fact stated in "Introduction to Coding Theory" by J. H. Van Lint, according to this book it follows from the transitiviness of the uniqueness of Extended Golay Binary Code ($G_{24}$) added to the fact that $Aut(G_{24})$ is transitive. Also I am not truly convinced with the transitive part (the uniqueness of $G_{24}$ part is proved and in my search on the internet I have found several proofs so I am fine with that) as it says it follows from the various constructions of $G_{24}$ (to keep it short several are explained in wikipedia).
As it is encouraged that in your question you should give your thoughts I will give my thoughts so far (in an informal manner): for each code $C$ with the parameters of $G_{23}$ you can add a parity number so the extended code is $G_{24}$ (or equivalent), as $G_{24}$ is unique and you can get $G_{23}$ by taking one component (I think that in this parts we invoque transitivity to say that the component we decide to eliminate does not matter), then $C$ would need to be an equivalent code to $G_{23}$.
 A: Let $C_1$ and $C_2$ be two perfect binary codes of length $23$, rank $12$ and minimum distance $7$. As you observed, extending either of them with an overall parity check bit gives you a code that is equivalent to $G_{24}$. Denote these two extended codes $C_1^+$ and $C_2^+$.
Let's fix a copy $G$ of $G_{24}$. We know that $C_1^+$ is equivalent with it, so there exists a coordinate permutation $\alpha\in S_{24}$ such that $\alpha(C_1^+)=G$. We don't know where the extension bit of $C_1^+$ went in this permutation, so let's call its new position $i=\alpha(24)$. Similarly, we see that there exists a permutation of bit positions $\beta\in S_{24}$ such that $\beta(C_2^+)=G$. The other extension bit was mapped to position $j=\beta(24)$.
By transitivity of the automorphism group of $G$ there exists an automorphism $\sigma$ of the code $G$ such that $\sigma(i)=j$. It follows that the composed permutation of bit positions, $\beta^{-1}\sigma\alpha$ takes the extension bit of $C_1^+$ to the extension bit of $C_2^+$, and hence gives the desired equivalence between the codes $C_1$ and $C_2$.
Essentially we get $C_1$ by puncturing the $i$th bit from $G$ and $C_2$ by puncturing the $j$th bit from $G$. The automorphism $\sigma$ then takes one punctured code to the other.
