Let $p\in [1,\infty)$. Given a $3\times 2$ matrix $A$, let $L_A:(\mathbb{R}^2, \| \cdot \|_p) \to (\mathbb{R}^3, \| \cdot \|_p )$ be an operator defined by $L_A(z) = Az$.
Question: If $L_A$ is an isometry, does it mean that the set of columns of $A$ form an orthonormal set?
Since $L_A$ is an isometry, by considering $L_A(1,0)^T$ and $L_A(0,1)^T$, clearly the set of columns of $A$ has norm $1$. However, I am not able to prove that they are orthogonal. Clearly the question has positive answer if $p=2$ as one can use inner product induced by the norm $\|\cdot\|_2$ to show that the columns of $A$ are orthogonal. It would be good if there is a reference for this fact, if it is true.
Remark: Let $(a_{11},a_{21}, a_{31})^T, (a_{12},a_{22}, a_{32})^T$ be the two columns of $A$. We say that $\{(a_{11},a_{21}, a_{31})^T, (a_{12},a_{22}, a_{32})^T\}$ is orthonormal if $\|(a_{11},a_{21}, a_{31})\|_p = \|(a_{12},a_{22}, a_{32})\|_p = 1$ and $a_{11}a_{12} + a_{21}a_{22} + a_{31}a_{32}= 0$.