$\ell^p$ is not isometric to $\ell^q$ The problem is this: if $1\le p<q<\infty$ then $\ell^p$ and $\ell^q$ are not isometric (as Banach spaces).
This is an exercise but I'd like to see an elegant proof.
 A: Let me complete the accepted answer by showing that $\ell^p$ is not isometric to $\ell^q$ if $1<p<2$ and $2<q<\infty$. Fix $1<r<\infty$. I claim that

Claim 5' The second derivative $\frac{d^2}{dt^2}\|x+ty\|_{\ell^r}\Big|_{t=0}$
  exists for every $x\in\ell^r\setminus\{0\}$ and every $y\in\ell^r$ if and only if $r\ge 2$.

Clearly, the claim implies the first assertion and thus finishes the problem.
Proof If $r<2$ then we take $x:=e_1$, $y:=e_2$ and we see that the second derivative $\frac{d^2}{dt^2}(1+|t|^r)^{1/r}\Big|_{t=0}$ does not exist. On the other hand, let $r\ge 2$: we claim that
$$\begin{aligned}\frac{d}{dt}\|x+ty\|_{\ell^r}\Big|_{t=0}&=\sum_{n=1}^\infty x'_ny_n=:A(x,y), \\ \frac{d^2}{dt^2}\|x+ty\|_{\ell^r}\Big|_{t=0}&=(r-1)\sum_{n=1}^\infty x''_ny_n^2-\frac{(r-1)}{\|x\|}\Big(\sum_{n=1}^\infty x'_ny_n\Big)^2=:B(x,y),\end{aligned}$$
where $x'\in\ell^{r/(r-1)}$ and $x''\in\ell^{r/(r-2)}$ are defined by
$$x'_n:=\frac{|x_n|^{r-2}x_n}{\|x\|_{\ell^r}^{r-1}},\quad x''_n:=\frac{|x_n|^{r-2}}{\|x\|_{\ell^r}^{r-1}}.$$
Notice that $x'$ and $x''$ depend continuously on $x\neq 0$ (which is easily seen by dominated convergence) and $\|x'\|_{\ell^{r/(r-1)}}=1$, $\|x''\|_{\ell^{r/(r-2)}}=\|x\|_{\ell^r}^{-1}$. So, by the generalized Holder's inequality, $A(x,y)$ and $B(x,y)$ are continuous and satisfy $|A(x,y)|\le\|y\|_{\ell^r}$, $|B(x,y)|\le 2(r-1)\|x\|_{\ell^r}^{-1}\|y\|_{\ell^r}^2$.
These formulas are a straightforward computation if $x,y\in c_c$ ($c_c$ being the space of sequences where only finitely many terms are nonzero). In order to check that they are true in general, we observe that for $x,y\in c_c$
$$\|x+ty\|_{\ell^r}=\|x\|_{\ell^r}+tA(x,y)+\int_0^t(t-s)B(x+sy,y)\,ds$$
(Taylor's expansion with remainder in integral form) and, by a density argument and dominated convergence, we see that this equation holds for all $x\in\ell^r\setminus\{0\}$ and all $y\in\ell^r$. Hence,
$$\begin{aligned}\frac{d}{dt}\|x+ty\|_{\ell^r}&=A(x,y)+\int_0^t B(x+sy,y)\,ds, \\ \frac{d^2}{dt^2}\|x+ty\|_{\ell^r}\Big|_{t=0}&=B(x,y).\end{aligned}$$
This finishes the proof. $\square$
A: Remark   As is well-known, the Banach spaces $c_0$ and $\ell^p$ for $1\leq p<\infty$ are mutually non-isomorphic, as a consequence of Pitt's theorem. See Corollary 2.1.6 in Albiac-Kalton's book. It seems a bit easier to prove the weaker result saying that the $\ell^p$ spaces are mutually non-isometric.
Note that $\ell^1$ is not strictly convex, while Clarkson's inequalities show that $\ell^p$ is strictly convex for every $1<p<\infty$. So $\ell^1$ is not isometric to $\ell^p$ for any $1<p<\infty$. We could obtain the same conclusion using reflexivity or uniform convexity instead of strict convexity.

Now assume there exist $1<p,q<\infty$ such that $\ell^p$ and $\ell^q$ are isometric via some isometry $T$: $\|Tx\|_p=\|x\|_q$ for every $x\in \ell^q$. We need to show that $p=q$. 

Sketch  An elementary manipulation of Clarkson's inequalities (see here for a short proof by R.P. Boas) yields Claim 1 below. Claims 2,3,4 follow immediately. This allows us to rule out every case but $q=p$ and $q=p'$. So it only remains to check that $\ell^2$ is the only $\ell^p$ space which is isometric to its dual (i.e. Claim 5, essentially).
We will denote $p'$ the conjugate exponent associated with $p$, i.e. $\frac{1}{p}+\frac{1}{p'}=1 \iff p'=\frac{p}{p-1}$. Recall that the continuous dual $(\ell^p)'$ is isometric to $\ell^{p'}$ for every $1<p<\infty$.

Claim 1   If $p\geq 2$, then $p'\leq q\leq p$.

Proof   For every $x,y \in \ell^q$, we have
$$
\big\| \frac{x+y}{2} \big\|_q^p+\big\| \frac{x-y}{2} \big\|_q^p=\big\| \frac{Tx+Ty}{2} \big\|_p^p+\big\| \frac{Tx-Ty}{2} \big\|_p^p
$$
$$
\leq \frac{\|Tx\|_p^p+\|Ty\|_p^p}{2}=\frac{\|x\|_q^p+\|y\|_q^p}{2}
$$
where we used Clarkson's inequality for $p\geq 2$. In particular, for $x=(1,1,0,\ldots)$ and $y=(1,-1,0,\ldots)$, this yields 
$$2=1+1\leq \frac{2^\frac{p}{q}+2^\frac{p}{q}}{2}=2^\frac{p}{q}\quad\Rightarrow\quad q\leq p$$
While for $x=(2,0,\ldots)$ and $y=(0,2,0,\ldots)$, we get
$$
2\cdot 2^\frac{p}{q}=2^\frac{p}{q}+2^\frac{p}{q} \leq \frac{2^p+2^p}{2}=2^p \quad\Rightarrow\quad p'=\frac{p}{p-1}\leq q
$$
Therefore we have $p'\leq q\leq p$. $\Box$

Claim 2   If $p\leq 2$, then $p\leq q\leq p'$.

Proof   Taking the dual,  $\ell^p\simeq \ell^q$ yields an isometric isomorphism $\ell^{p'}\simeq \ell^{q'}$. Since $p'\geq 2$, Claim 1 applied to $p'$ implies $(p')'\leq q'\leq p'\iff p\leq q\leq p'$. $\Box$

Claim 3   If $p$ and $q$ are both $\geq 2$ or both $\leq 2$, then $p=q$.

Proof   Assume $p,q\geq 2$. Applying Claim 1 to $p$ yields $q\leq p$, while applying it to $q$ gives $p\leq q$. Hence $p=q$. The case $p,q \leq 2$ follows from Claim 2 in a similar manner. Or you can just take the duals and conclude that $p'=q'$ since they are both $\geq 2$. $\Box$

Claim 4   If $p\geq 2$ and $q\leq 2$, then $q=p'$.

Proof   By Claim 1 for $p$, we have $p'\leq q$. By Claim 2 for $q$, we get $p\leq q' \iff q\leq p'$. Hence $q=p'$. $\Box$

Claim 5   If $\ell^p$ is isometric to $\ell^{p'}$, then $p=p'=2$.

Proof   Still pondering what the easiest argument could be...
[Edit by the OP: see my answer below, which should prove Claim 5 and finish the problem off.]
A: I think it should work.
Suppose for definiteness $p < q$. Then $\ell^p \subset \ell^q$. Assume that an isometry exists $J: \ell^p \to \ell^q$. Then $J(\ell^p) \subset \ell^q$ and in particular $J(B_{\ell^p}) \subset B_{\ell^q}$, where $B_X$ is the unit ball of $X$. Let us choice $r >0$ s.t. $r B_{\ell^q} \subset J(B_{\ell^p})$ (such an $r$ must exist, since $J(B_{\ell^p})$ is nonempty). Being an isometry, $J$ is an isomorphism from $\ell^p$ onto $J(\ell^p)$, i.e. into $\ell_q$, so by Pitt's theorem it must be compact. But $J(B_{\ell^p})$ can't be compact, since $rB_{\ell^q}$ is not. Hence $J$ can't be an isometry.
Remark 1. Take with care! It seems to me that it works but wait for community peer-review!
Remark 2. It is well known that every Banach separable space is isometrically embedded into a closed subspace of $\ell^\infty$, so we are correctly considering only the case $p,q < \infty$. The proof of this fact relies on the Hahn-Banach theorem; maybe one can adapt it to our case.
A: WARNING: This answer is based on a (most probably) false assumption and so it is (most probably) wrong. See comments
Here's a partial answer which works when $p$ and $q$ are even integers. This extra assumption is used in the shaded area below.
If a linear isometry of $\ell^p$ onto $\ell^q$ existed, its restriction to the subspace 
$$\left\{ (x_1, x_2, 0, 0 \ldots)\ :\ x_1, x_2\in \mathbb{R}\right\}\subset \ell^p$$
would yield a linear isometry of $(\mathbb{R}^2, \lvert\cdot\rvert_p)$ onto $(\mathbb{R}^2, \lvert\cdot\rvert_q)^{[1]}$. Therefore we only need to show that the latter cannot exist. 

Since $p$ is an even integer, the $\lvert\cdot\rvert_p$-unit
  circle is an algebraic curve of degree $p$:  $$S_p=\left\{ (x_1, x_2)\
 :\ x_1^p+x_2^p=1\right\}, $$ and any nonsingular linear operation
  transforms it into an algebraic curve of the same degree. This rules
  out the existence of a linear isometry of $(\mathbb{R}^2,
 \lvert\cdot\rvert_p)$ onto $(\mathbb{R}^2, \lvert\cdot\rvert_q)$
  because such a mapping would transform $S_p$ into $S_q$ and the latter has degree $q$.

$^{[1]}$ Where $\lvert (x_1, x_2)\rvert_p=\left( \lvert x_1\rvert^p+\lvert x_2\rvert^p\right)^{\frac{1}{p}}.$
