The standard 3 code scheme encodes one qubit into 3 by applying 2 $CNOT$s targeted on auxiliary qubits set on ground state $|0\rangle$.
I am struggling to perform logical operations between two logical qubits encoded this way, so I am starting to suspect that the operation $H^{\otimes 3}$ is not a logical operation for such a code.
The logical computational basis states of the $3$-qubit repetition code are $|0_L\rangle=|000\rangle$ and $|1_L\rangle=|111\rangle$ which are unentangled. However, the logical Hadamard sends $|0_L\rangle$ to $|{+}_L\rangle=\frac{1}{\sqrt2}(|000\rangle+|111\rangle)$ which is entangled. Every transversal gate is a product of local unitaries $U_1\otimes U_2\otimes U_3$ which cannot create entanglement between the physical qubits. Therefore, there is no transversal Hadamard in the repetition code.
There's another method of reasoning this. Note that $H^{\otimes 3} |0_{L}\rangle$ creates the uniform superposition over all computational basis states.
Since the logical codespace is spanned by just $|000\rangle$ and $|111\rangle$, this is obviously not even a codestate, let alone the logical $|{+}_{L}\rangle$ state.
You can use this to immediately rule out many other potential 'logical' operations, too.
