An operator on the Hilbert space of $ n $ qubits is called a local unitary if it is of the form $$ U=\bigotimes_{i=1}^n g_i $$ where each $ g_i $ is a $ 2 \times 2 $ unitary matrix. In other words if it is an element of the group of local unitaries $$ \bigotimes_{i=1}^n U(2) $$ The group of all non-entangling gates on an $ n $ qubit Hilbert space is exactly the normalizer of the group of local unitaries $$ N \Bigg( \bigotimes_{i=1}^n U(2) \Bigg) \cong \bigotimes_{i=1}^n U(2) \rtimes S_n $$ The structure of this group as a subgroup of $ U(2^n) $ is a bit difficult to describe. Fortunately, in quantum mechanics global phase is irrelevant so we really work in a projective Hilbert space and the operations are projection unitaries. So modding everything out by a $ U_1 $ group of global phases the group of local projective unitaries as a subgroup of $ PU(2^n) $ becomes $$ \prod_{i=1}^n PU(2) $$ and it turns out the group of all non-entangling gates in $ PU(2^n) $ is exactly $$ N \Bigg(\prod_{i=1}^n PU(2) \Bigg)= \Bigg(\prod_{i=1}^n PU(2) \Bigg) \rtimes S_n $$ where the symmetric group on $ n $ letters acts by permuting the $ n $ many $ PU(2) $ groups in the direct product in the natural way.
$ 2 $ is not special in this discussion and can be replaced with any qudit dimension $ d $.
Now we define code equivalence, following Raines
https://arxiv.org/abs/quant-ph/9704043
Two codes are locally equivalent if they are related by a local unitary. And we will say two codes are globally equivalent, or just equivalent, if they are related by a local unitary and a permutation (in other words, they are related by a non-entangling gate).
Let $ C_1,C_2 $ be two codes with the same parameters $[\![n,k,d]\!]$. Is it true that that $ C_1 $ and $ C_2 $ are equivalent if and only if they have isomorphic transversal gate groups?
Note that two codes which are equivalent must have conjugate and thus isomorphic transversal gate group.
