If $(X,| \cdot|)$ is isometrically isomorphic with $(X,\|\cdot \|)$ is it always true that the norms are equivalent?? Let $(X,|\cdot|)$ and $(X,\| \cdot\|)$ be normed vector spaces and $T:X\to X$ an isometric isomorphism. Is it true that the norms $| \cdot |$ and $\|\cdot \|$ are equivalent? I think in general case that DOES not hold but I am not pretty sure. Any conditions to make this be true?
Edit: An idea is when $\dim X<+\infty$ where all norms are equivalent.
 A: If $X$ has two norms, $\|\cdot\|$ and $|\cdot|$, then:

*

*It is possible for the norms to be equivalent but the spaces are not isometrically isomorphic.
For example, let $X=\mathbb{R}^2$ with 1-norm and 2-norm. These norms are equivalent $$\|x\|_2\le\|x\|_1\le\sqrt{2}\|x\|_2$$ but only the 2-norm satisfies the parallelogram law. If $T:X_1\to X_2$ were an isometric isomorphism, $\|Tx\|_2=\|x\|_1$, then for $x=(1,1)$, $y=(1,-1)$, $$\|x+y\|_1^2+\|x-y\|_1^2=8<16=2\|x\|_1^2+2\|y\|_1^2$$ yet $$\|Tx+Ty\|_2^2+\|Tx-Ty\|_2^2=2\|Tx\|_2^2+2\|Ty\|_2^2$$ gives a contradiction.


*It is possible for the norms to be inequivalent but the spaces are isometrically isomorphic.
Let $X$ be a (infinite-dimensional) normed space. Take any bijective but unbounded linear map $T:X\to X$. Then it is trivial to check that $\|Tx\|$ is a norm; call it $|x|:=\|Tx\|$, so $T$ becomes an isometric isomorphism. But the norms are inequivalent, for $|x|\le c\|x\|$ would imply that $T$ is bounded$$\|Tx\|=|x|\le c\|x\|$$
For a specific example, take $X=c_{00}$ with the $\infty$-norm and $T(a_n)=(na_n)$. So $\|(a_n)\|=\max_n|a_n|$, $|(a_n)|=\max_nn|a_n|$. Then $X_{\|\cdot\|}$ is isometrically isomorphic to $X_{|\cdot|}$ via $T$, but the norms are not equivalent $$|e_n|=n\not\le c\|e_n\|$$
[In finite dimensions, any two norms are equivalent.]
