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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.

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    $\begingroup$ What does logical counterpart of $U$ mean? $\endgroup$
    – FDGod
    Mar 22 at 6:34
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    $\begingroup$ @FDGod I mean it's one of the logical $U$. I was not sure about the uniqueness of the logical $U$ so I tried to be cautious about my word-use. $\endgroup$
    – Yunzhe
    Mar 22 at 7:46

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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.

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