Example 3 Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}, a \neq 0)$ satisfy the conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and $f(x) \geqslant x$;
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.

Find the largest real number $m(m>1)$ such that there exists $t \in \mathbf{R}$, for which, if $x \in[1, m]$, then $f(x+t) \leqslant x$. (2002 National High School Mathematics Competition Problem)
From (1), we know that $x=-1$ is the axis of symmetry of $f(x)$, so $-\frac{b}{2a}=-1, b=2a$. Also, $f(1) \geqslant 1$. From (2), we know that $f(1) \leqslant 1$, hence $f(1)=1, a+b+c=1$. Since $f(-1)=0, a-b+c=0$. Therefore, $a=\frac{1}{4}, b=\frac{1}{2}, c=\frac{1}{4}$. Thus, $f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x+\frac{1}{4}=\frac{1}{4}(x+1)^{2}$.
If there exists $t \in \mathbf{R}$, such that for any $x \in [1, m]$, we have $f(x+t) \leqslant x$. Let $x=1$, then $f(t+1) \leqslant 1$, which means $\frac{1}{4}(t+2)^{2} \leqslant 1$. Solving this, we get $-4 \leqslant t \leqslant 0$.
For a fixed $t \in [-4,0]$, let $x=m$, then $f(t+m) \leqslant m$.
Simplifying, we get
$$
m^{2}-2(1-t) m+\left(t^{2}+2 t+1\right) \leqslant 0
$$

Solving this, we get
$$
1-t-\sqrt{-4 t} \leqslant m \leqslant 1-t+\sqrt{-4 t}
$$

Thus, $\quad m \leqslant 1-t+\sqrt{-4 t} \leqslant 1-(-4)+\sqrt{(-4) \cdot(-4)}=9$
On the other hand, when $t=-4$, for any $x \in [1,9]$, we always have
$$
f(x-4)-x=\frac{1}{4}\left(x^{2}-10 x+9\right)=\frac{1}{4}(x-1)(x-9) \leqslant 0
$$

In summary, the maximum value of $m$ is 9.