15. Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}, a \neq 0)$ satisfy the following 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 $m(m>1)$ such that there exists $t \in \mathbf{R}$, for any $x \in[1, m]$, $f(x+t) \leqslant x$.
15. Since $f(x-4)=f(2-x)$, the graph of the function is symmetric about $x=-1$,

so $-\frac{b}{2 a}=-1, b=2 a$.
From (3) when $x=-1$, $y=0$, i.e., $a-b+c=0$,
From (1) we get $f(1) \geqslant 1$, from (2) we get $f(1) \leqslant 1$,
so $f(1)=1$, i.e., $a+b+c=1$, and $a-b+c=0$,
so $b=\frac{1}{2}, a=\frac{1}{4}, c=\frac{1}{4}$, thus $f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x+\frac{1}{4}$.
Assume there exists $t \in \mathbf{R}$, such that for any $x \in[1, m]$, $f(x+t) \leqslant x$.
Taking $x=1$, we have $f(t+1) \leqslant 1$. That is, $\frac{1}{4}(t+1)^{2}+\frac{1}{2}(t+1)+\frac{1}{4} \leqslant 1$, solving this gives $-4 \leqslant t \leqslant 0$.
For a fixed $t \in[-4,0]$, taking $x=m$, we have $f(t+m) \leqslant m$, i.e., $\frac{1}{4}(t+m)^{2}+\frac{1}{2}(t+m)+\frac{1}{4} \leqslant m$.
Simplifying, we get
$$
m^2 - 2(1-t) m + (t^2 + 2t + 1) \leqslant 0,
$$

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

Thus, we have $\quad m \leqslant 1-t+\sqrt{-4t} \leqslant 1-(-4)+\sqrt{-4(-4)}=9$.
When $t=-4$, for any $x \in[1,9]$, we always have
$$
f(x-4)-x=\frac{1}{4}(x^2-10x+9)=\frac{1}{4}(x-1)(x-9) \leqslant 0 .
$$

Therefore, the maximum value of $m$ is 9.