Suppose we have a $[[n, 1]]$ stabilizer code $Q$ and a single-qubit unitary $U$. We define the logical counterpart of $U$ as $\bar{U}$. My question is: Is there just one $\bar{U}$ up to stabilizers of $Q$?
I am asking this because I have seen the definition of transversal gate to be like $\bar{U}=U^{\otimes n}$, while the right side seems to be unique if $U$ is given.
The answer turns out to be NO as denoted in this answer. There seems to be infinite number of operations that preserves the codespace of $Q$ and acts as if $\bar{U}$ for any given single-qubit unitary $U$.
As an example, we can define $Q$ to be a two-qubit trivial code with stabilizer $IZ$ and logical operator $X_L = XI, Z_L=ZI$. If $U=H$ is a Hadamard gate, then any operations with the following form $$ H\otimes |0\rangle\langle0| + A\otimes |1\rangle\langle1| $$ is a logical $\bar{H}$, where $A$ is an arbitrary single-qubit unitary.