6. $65 X=\{0, a, b, c\}$ and $M(X)=\{f \mid f: X \rightarrow X\}$ is the set of all functions from $X$ to itself. Define the addition operation $\oplus$ on $X$ as follows:
\begin{tabular}{|c|c|c|c|c|}
\hline$\oplus$ & 0 & $a$ & $b$ & $c$ \\
\hline 0 & 0 & $a$ & $b$ & $c$ \\
\hline$a$ & $a$ & 0 & $c$ & $b$ \\
\hline$b$ & $b$ & $c$ & 0 & $a$ \\
\hline$c$ & $c$ & $b$ & $a$ & 0 \\
\hline
\end{tabular}
(1) If
$$
\begin{aligned}
S & =\{f \in M(X) \mid f((x \oplus y) \oplus x)=(f(x) \oplus f(y)) \oplus f(x), \\
\forall x, y & \in X\},
\end{aligned}
$$

determine the number of elements in $S$.
(2) If
$$
I=\{f \in M(X) \mid f(x \oplus x)=f(x) \oplus f(x), \forall x \in X\},
$$

determine the number of elements in $I$.
[Solution] (1) From the addition operation table, for $x \in \{a, b, c\}$,
we have
$$
\begin{array}{l}
0 \oplus x = x \oplus 0 = x, \\
x \oplus x = 0, \\
b \oplus a = a \oplus b = c, \quad c \oplus a = a \oplus c = b, \\
c \oplus b = b \oplus c = a.
\end{array}
$$

Therefore, we have
$$
x \oplus y = y \oplus x, \forall x, y \in X.
$$

And it is easy to verify that
$$
(x \oplus y) \oplus x = y, \forall x, y \in X.
$$

In fact, if $y = 0$, then
$(x \oplus y) \oplus x = (x \oplus 0) \oplus x = x \oplus x = 0 = y$.
If $y \neq 0$, but $x = 0$, then
$(x \oplus y) \oplus x = (0 \oplus y) \oplus 0 = y \oplus 0 = y$.
If $y \neq 0$ and $x \neq 0$, let $z = \{a, b, c\} - \{x, y\}$, then $(x \oplus y) \oplus x = z \oplus x = y$.
Therefore, (1) holds.
Since $\forall x, y \in X$, we have $f(x), f(y) \in X$, so by (1) we get
$$
(f(x) \oplus f(y)) \oplus f(x) = f(y).
$$

Therefore,
$$
S = M(X).
$$

But
$$
|M(X)| = 4^4 = 256,
$$

so
$$
|S| = 256,
$$

i.e., the number of elements in $S$ is 256.
(2) Since
$$
\begin{array}{l}
f(x \oplus x) = f(0), \\
f(x) \oplus f(x) = 0,
\end{array}
$$

we have
$$
I = \{f \in M(X) \mid f(0) = 0\},
$$

thus
$$
|I| = 4^3 = 64,
$$

i.e., the number of elements in $I$ is 64.