$(\mathbb{R}^n,\|\cdot\|_{p})$ is isometrically isomorphic to $(\mathbb{R}^n,\|\cdot\|_{q})$ iff $p=q,$ I want to prove, that $(\mathbb{R}^n,\|\cdot\|_{p})$ is isometrically isomorphic to $(\mathbb{R}^n,\|\cdot\|_{q})$ iff $p=q,$ I have tried looking the unitary balls, and I have been proved that $(\mathbb{R}^n,\|\cdot\|_{1})$ is not isometrically isometric to any other $p$-norm.
 A: In the article Isometries of Finite-Dimensional Normed Spaces by F. C. Sanchez and J.M.F. Castillo, the authors prove the following result.

Theorem. For $1\le p,q\le\infty$ and $n\ge 2$, $(\mathbb{R}^n,\|\cdot\|_p)$ and $(\mathbb{R}^n,\|\cdot\|_q)$ are isometrically isomorphic if and only if $p=q$ or if $n=2$ and $p,q\in\{1,\infty\}$.

The argument splits in several cases.
The first one is to rule out any isometric isomorphisms between $\ell^n_p$ and $\ell^n_q$ if $q\not=p'$ where $p'$ is the Hölder dual of $p$ defined by $\frac1p+\frac1{p'}=1$.
This is a very beautiful argument, which I will now try to reproduce. For the remaining cases, see the paper.
For simplicity let us also only consider $1<p,q<\infty$.
Denote $(\mathbb{R}^n,\|\cdot\|_p)$ by $\ell^n_p$.
Say $\phi:\ell^n_p\to \ell^n_q$ is an isometric isomorphism.
Recall that, by Hölder's inequality (and its condition for equality), the dual space of $\ell^n_p$ is given by $(\ell^n_p)^*=\ell^{n}_{p'}$ (up to isometric isomorphism). 
The dual map $\phi^*:\ell_{q'}^n\to \ell_{p'}^n$ is also an isometric isomorphism (this is a general fact, see appendix below). Unpacking definitions, we see that there exists an invertible matrix $A\in\mathbb{R}^{n\times n}$ such that
$$ \|Ax\|_q = \|x\|_p$$
and
$$ \|A^* x\|_{p'}=\|x\|_{q'}$$
hold for all $x\in\mathbb{R}^n$, where $A^*$ is the dual matrix (so the transpose) of $A$.
The question is: How does such a matrix look like?
Observe that the unit vectors have norm $1$ in all the $p$-norms. So setting $x=e_j$ in both equations we get
$$1=\|Ae_j\|^q_q = \sum_{i=1}^n |A_{ij}|^q$$
and
$$1=\|A^T e_j\|^{p'}_{p'}=\sum_{i=1}^n |A_{ji}|^{p'}$$
Adding both equations over $j=1,\dots,n$ we obtain
$$n=\sum_{i,j=1}^n |A_{ij}|^q = \sum_{i,j=1}^n |A_{ij}|^{p'},$$
so the entry-wise $q$ norm of $A$ equals the entry-wise $p'$ norm and both equal $n$.
Observing also that $|A_{ij}|\le 1$ (by the previous) and since $q\not=p'$ we see that each  $A_{ij}$ can only be $0$ or $\pm 1$ and the number of non-zero entries is exactly $n$. Since the matrix is also invertible, we now know exactly how it looks like: it has exactly one $\pm 1$ in each row and each column. That is, $A$ is a "generalized" permutation matrix (the "generalized" hinting to the fact that also $-1$ is allowed). 
(the beauty is that we have now "accidentally" classified all the isometries $\ell^n_p\to \ell^n_p$ if $p\not=2$!, which would have been an interesting follow-up question)
From here it is easy to conclude that we must have $p=q$. Consider the vector $x=e_1+e_2$. Then $Ax=\epsilon_1 e_i + \epsilon_2 e_j$ for some $i,j$ and choices of signs $\epsilon_1,\epsilon_2\in\{\pm 1\}$, so 
$$2^{1/p}=\|x\|_p=\|Ax\|_q = 2^{1/q}.$$
So $p=q$.

Appendix.

Lemma. Let $X,Y$ be normed vector spaces and $\phi:X\to Y$ an isometric isomorphism. Then also the dual map
  $$\phi^*:Y^*\to X^*,$$
  given by $\phi^*(w)(x)=w(\phi(x))$ is an isometric isomorphism.

Proof. 
We have $$\|\phi^* w\|_{X^*}=\sup_{x\not=0} \frac{|w(\phi(x))|}{\|x\|_X}=\sup_{x\not=0} \frac{|w(\phi(x))|}{\|\phi(x)\|_Y} = \sup_{y\not=0} \frac{|w(y)|}{\|y\|_Y}=\|w\|_{Y^*},$$
where the penultimate step uses that $\phi$ is bijective.
$\square$
