= Solution
Suppose a bounded linear isomorphism $T:\ell^p\to Y\subset\ell^2$ with bounded inverse existed. There would be constants $0<c\leq C$ such that $c\|x\|_p\leq\|Tx\|_2\leq C\|x\|_p$. Put $y_j=Te_j$. Averaging the squared Hilbert norm over all $2^n$ independent sign choices cancels cross terms, giving the <generalized parallelogram identity>
$$
2^{-n}\sum_{\varepsilon_j=\pm1}\left\|\sum_{j=1}^n\varepsilon_jy_j\right\|_2^2
=\sum_{j=1}^n\|y_j\|_2^2.
$$
Every signed sum of the original $e_j$ has $p$-norm $n^{1/p}$. The left side is therefore between $c^2n^{2/p}$ and $C^2n^{2/p}$, while the right side is between $c^2n$ and $C^2n$. Thus $n^{2/p}$ and $n$ would be bounded constant multiples of one another for all $n$, impossible unless $p=2$. Consequently \b[no such normed-space isomorphism exists for $p\ne2$].
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