Solution:

Let $k$ be a snug integer. Note that any factor that divides $n$, $k$, and $k+1$ must also divide $(k+1)-k=1$, so
$$\gcd(k, n) = \gcd(k+1, n) = 1.$$
In particular, $k$ has a multiplicative inverse $h \bmod n$ (we can choose $h$ such that $0 < h < n$). We claim that $h$ is also snug. Clearly $\gcd(h, n) = 1$; note that
$$(h+1) \cdot k = h k + k \equiv k + 1 \pmod{n};$$
since $k$ and $k+1$ are invertible $\bmod n$, so is $h+1$.
Thus we can pair up snug residues $\bmod n$ into pairs with product $1$, unless there is a snug $k$ that is its own multiplicative inverse. We claim that there is no such $k$ except possibly $k=1$. Indeed, if $k^2 \equiv 1 \pmod{n}$, then $n$ divides
$$k^2 - 1 = (k+1)(k-1).$$
Since $k$ is snug, $n$ is relatively prime to $k+1$ and hence divides $k-1$, implying that $k=1$.