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I wanted to work through the details in the main answer to check I understood the conditions for the existence of logical Hadamard gates. But I am slightly confused as to whether we are implicitly assuming that the code has only 1 logical qubit.
So, let $C \leq \mathbb{F}_2^n$ be a classical code such that $C\leq C^{\perp}$. It's well known that we can construct a "self-dual" CSS code from $C$, where our stabilisers are $\{X^v, Z^v:v \in C\}$.
Let's pick bases for the classical spaces $$ \langle c_1,\dots,c_k \rangle =C \leq C^{\perp} = \langle c_1,\dots,c_k, d_1,\dots, d_l \rangle $$
In particular our CSS code has $l = \dim (C^{\perp} \backslash C)$ logical qubits. Furthermore, there exists another basis of $C^{\perp} \backslash C$, say $\{e_1,\dots,e_l\}$ such that we may use the two bases to define our logical Pauli ops: $$\bar{X_i}=X^{d_i}, \bar{Z_i}=Z^{e_i}$$
Now, my understanding of transversality is that we want to apply an operator $U$ to every physical qubit and have this implement logical $\bar{U}$ on each encoded bit.
So for $H^n$ to implement $\bar{H}$ on each logical qubit, it must conjugate the encoded Paulis appropriately. This holds iff $d_i = e_i$ and furthermore we require $e_i\cdot e_j=\delta_{ij}.$
This is where I struggle, as we are essentially asking for an orthogonal basis of a binary space, which isn't true in general. NB: it IS true if $l=1$ i.e. we have exactly one logical qubit
An example: Consider the following classical codes $C\leq C^{\perp}$. $$ C = \langle (111001), (000110) \rangle$$ $$ C^{\perp} = \langle (111001), (000110), (010001), (001001) \rangle $$.
This encodes a CSS code with 2 logical qubits but note that the space $\langle (010001), (001001) \rangle$ contains only vectors of even weight.
Here the only choice (up to stabilisers) of logical ops are $$ \bar{X_1}=X^{010001}, \bar{X_2}=X^{001001} $$ $$ \bar{Z_1}=Z^{001001}, \bar{Z_2}=Z^{010001} $$
So in particular $H^n$ actually sends $\bar{X_1}$ to $\bar{Z_2}$ and therefore isn't logical hadamard. In fact $H^n$ corresponds to logical Hadamard $H^2$ combined with a permutation of the logical qubits.
It'd be great if someone could clarify this for me -- am I misunderstanding the notion of transversality on multi-logical qubit codes?