Given any $m\times m$ square matrix $M$, let $L_M:(\mathbb{R}^m, \|\cdot\|_p) \to (\mathbb{R}^m, \|\cdot\|_q)$ be defined by $L_M(x) = Mx$ where $1\leq p,q< \infty$ and $$\|(x_1,...,x_m)\|_p= \left( \sum_{k=1}^m |x_k|^p \right)^{1/p}.$$
It is well-known that if $p=q=2$ and $L_M$ is an isometry, then the matrix $M$ is orthogonal. I am interested to know whether the same holds if we assume that $p\neq q$. More precisely,
Assume that $L_M$ is an isometry. If $p\neq q$, must $M$ be an orthogonal matrix?
Clearly each column of $M$ has norm $1$. So, it remains to show that every two columns of $M$ has zero dot product. But I got stuck here.
The unit spheres in $\mathbb R^m$ with respect to $\|\cdot\|_p$ and $\|\cdot\|_q$ aren't linear images of each other when $p\neq q$, hence there are no linear maps preserving norms in this case.
So the answer is trivially "yes": Every linear map $(\mathbb{R}^m, \|\cdot\|_p) \to (\mathbb{R}^m, \|\cdot\|_q)$ preserving norms is also orthogonal (since there are none).