I'll write $\mathbf{Z}^n$ for the integral quadratic form $x_1^2 + \cdots + x_n^2$. For which values of $n$ is $\mathbf{Z}^n$ unique in its genus, i.e. isolated in Kneser's graph? In particular can someone verify or shoot down the following guesses:
For $n \leq 8$, any lattice in the same genus as $\mathbf{Z}^n$ is isomorphic to $\mathbf{Z}^n$.
Any lattice in the same genus as $\mathbf{Z}^9$ is isomorphic to either $\mathbf{Z}^9$ or to $\mathbf{Z} \times \mathrm{E}_8$.
[edited mostly to add information about $n > 9$]
David Treumann's guess is correct: ${\bf Z}^n$ is unique in its genus iff $n \leq 8$, and for $n = 9$ the genus consists of only ${\bf Z}^9$ and ${\bf Z} \oplus E_8$.
The comments indicate two ways to prove this, using $p$-neighbors (as implemented in MAGMA) or the mass formula. Alternatively, one can use the fact that the theta function $\theta_L$ of any lattice $L$ in the genus of ${\bf Z}^n$ is a modular form of weight $n/2$ for an index-3 subgroup $\Gamma$ of the full modular group ${\rm PSL}_2({\bf Z})$.
For $n < 8$, there is only one choice of $\theta_L$ that has $q^0$ coefficient $1$, and we find that $L$ must have $2n$ vectors of norm $1$, and is thus isomorphic with ${\bf Z}^n$.
For $n=8$, either there are $16$ vectors of norm $1$ or there is a characteristic vector of norm zero. But in the latter case $L$ is an even lattice, and thus not in the genus of ${\bf Z}^8$. So we're back to having enough unit vectors to identify $L$ with ${\bf Z}^8$.
The case of $n=9$ requires a bit more work. We can always write $L = {\bf Z}^m \oplus L_0$ where $L_0$ is a unimodular lattice of rank $n-m$ with no vectors of norm $1$. Using $\theta_{L_0}$ we soon find that $L_0$ is either $E_8$ or the trivial lattice of rank zero. Hence $L$ is either ${\bf Z} \oplus E_8$ or ${\bf Z}^9$.
For the details, see for example my papers
A characterization of the ${\bf Z}^n$ lattice, Math. Research Letters 2 (1995), 321-326 (arXiv: math.NT/9906019).
Lattices and codes with long shadows, Math. Research Letters 2 (1995), 643-651 (arXiv: math.NT/9906086).
Of course for every $n \geq 10$ the genus still contains at least the two lattices ${\bf Z}^n$ and ${\bf Z}^{n-8} \oplus E_8$, so $n=8$ is the last case where ${\bf Z}^n$ is unique in its genus.
Added later: The genus of ${\bf Z}^n$ is known at least for $n \leq 25$. According to Table 2.2 on page 49 of SPLAG =
John Conway and Neil Sloane: Sphere Packings, Lattices and Groups, 3rd ed. New York: Springer 1999.
the size of this genus for $9 \leq n \leq 25$ is as follows: $$ \begin{array}{c|cccccccccccccc} n & 9,10,11 & 12,13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 \cr\hline \# & 2 & 3 & 4 & 5 & 6 & 9 & 13 & 16 & 28 & 40 & 68 & 117 & 273 & 665 \end{array} $$ Look up "Lattice, unimodular" in the index (page 694) for pointers in the book for further information.
Both statements are well-known in the arithmetic theory of quadratic forms (aka integral lattices).
The first statement for $n \leq 5$ is a consequence of Hermite's bound on the nonzero minimum of an integral lattice: the minimum of a unimodular lattice of rank $\leq 5$ is 1.
Both statements are consequences of Kneser's theory of neighbors. You can find it in the last theorem in O'Meara's book. It also mentioned that Kneser completed the analysis of the genus of $I_n$ for $n \leq 13$.
