10. Define $S_{n}=\{0,1, \cdots, n-1\}\left(n \in \mathbf{Z}_{+}\right)$.

For a bijection $f: S_{n} \rightarrow S_{n}$, if $n \mid(a+b-c), a, b, c \in S_{n}$, then
$$
n \mid(f(a)+f(b)-f(c)) \text {. }
$$

Let the set of all solutions of $f$ be $T_{n}$. If $\left|T_{n}\right|=60$, find $n$.
10. Take any $f \in T_{n}$, let $k=f(1)$.

From the given conditions, we have
$$
\begin{aligned}
& n \mid(f(0)+f(0)-f(0)) \\
\Rightarrow & n \mid f(0) \\
\Rightarrow & f(0)=0 \\
\Rightarrow & n \mid(f(m-1)+f(1)-f(m)) \\
& (1 \leqslant m \leqslant n-1) \\
\Rightarrow & f(m) \equiv k m(\bmod n)\left(m \in S_{n}\right) .
\end{aligned}
$$

In particular, $f$ depends on the value of $f(1)$.
Since $f$ is a bijection, we have
$$
\operatorname{gcd}(k, n)=1 \text {. }
$$

Thus, $\left|T_{n}\right| \leqslant \varphi(n)$.
For any $k \in S_{n}$, and $\operatorname{gcd}(k, n)=1$, the function satisfying
$$
f(m) \equiv k m(\bmod n)\left(m \in S_{n}\right)
$$

is a bijection, because the set $\{0, k, 2 k, \cdots, (n-1) k\}$ is a complete residue system modulo $n$.
Therefore, if $n \mid(a+b-c)$, then
$$
\begin{array}{l}
f(a)+f(b)-f(c) \equiv k(a+b-c) \\
\equiv 0(\bmod n) \\
\Rightarrow f_{k} \in T_{n} \\
\Rightarrow\left|T_{n}\right| \geqslant \varphi(n) .
\end{array}
$$

Combining (1) and (2), we get
$$
\left|T_{n}\right|=\varphi(n)(n>1) \text {. }
$$

Given $\varphi(n)=60$, we solve to get
$$
n=61,77,93,99,122,124,154,186,198 \text {. }
$$