If two norm $1$ vectors have the same $L1$ norm, are they the "same" vector?

Question

Suppose I have two vectors $u, v \in \mathbb{R}^n$ where $\|u\|_2 = \|v\|_2 = 1$. I am trying to see if these two vectors are the "same" under the following definition.

Two vectors are the "same" if you can change the sign, or shuffle the entries to make them equivalent.

For example, $[1,2]$ is the "same" as $[-1,2],[1,-2],[-1,-2],[2,1],[-2,1],[2,-1],[-2,-1]$. This is not a norm-$1$ vector but I just thought this was an easy example.

Now suppose $u,v$ are norm-$1$ such that $$ \sum u_i^2 = \sum v_i^2 = 1 $$ and that their associated L$1$ norms are equal $$ \sum |u_i| = \sum |v_i|. $$

Does that mean that the two vectors must be the "same"?

In $\mathbb{R}^2$ this seems to be true just by plotting the vectors $[\sin\theta, \cos\theta]^T$, and could probably be proven pretty easily using triginometry identities. However I was wondering if anyone knew if this generalized to higher dimensions $n$. Thank you for the help!

How about $(1,0)$ and $(\frac1 {\sqrt2} , \frac1 {\sqrt 2})$? — Kavi Rama Murthy, Dec 19 '20 at 23:49
In $\mathbb{R}^2$ the set of vectors with $L_2$-norm 1 is a sphere, and the vectors with a fixed $L_1$ norm is a plane (restricting to all positive entries for simplicity.) So imagine intersecting a plane and a sphere... — Jair Taylor, Dec 19 '20 at 23:52
@JairTaylor ah I see, in the next dimension you will have equivilent norm $1$ vectors that are not same to shuffling due to it being a continuous circle. That makes sense, if you make that an answer I would mark that correct — wjmccann, Dec 20 '20 at 00:02
I would generalise — same $1$-norm and same $2$-norm, where the $2$-norm is not necessarily $1$. Take a look at these hats. — Rodrigo de Azevedo, Dec 20 '20 at 00:04
@RodrigodeAzevedo ok so for an $n$ dimensional system, if the first $n$ norms are equal, then the two vectors are the "same"? — wjmccann, Dec 20 '20 at 00:07
My advice: rephrase your question to focus on the first $n$ norms. Then focus on the case $n=3$ and give it a try. Take a look at this. — Rodrigo de Azevedo, Dec 20 '20 at 00:12

score 2 · Accepted Answer · answered Dec 20 '20 at 00:11

Yes, in $\mathbb{R}^2$ (as well as, trivially, in $\mathbb{R}$); no, in $\mathbb{R}^n$ for $n\geq 3$.

The "yes in $\mathbb{R}^2$" part is easily seen by, as you said, considering the vectors $\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}$.

A counterexample to the statement in general would be the two vectors in $\mathbb{R^3}$, $\begin{bmatrix}-1/3\\2/3\\2/3\end{bmatrix}$ and $\begin{bmatrix}1\\0\\0\end{bmatrix}$.

If two norm $1$ vectors have the same $L1$ norm, are they the "same" vector?

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