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 which, if $x \in[1, m]$, then $f(x+t) \leqslant x$.
The function $f(x)$ is symmetric about the line $x=-1$, and $1 \leqslant f(1) \leqslant 1 \Rightarrow f(1)=1$, then $f(x)=a(x+1)^{2} \Rightarrow f(1)=4 a=1 \Rightarrow a=\frac{1}{4} \Rightarrow f(x)=\frac{(x+1)^{2}}{4}$.
By the AM-GM inequality, $f(x)=\frac{(x+1)^{2}}{4}=\frac{x^{2}+1+2 x}{4} \geqslant x$, thus $f(x)=\frac{(x+1)^{2}}{4}$ satisfies all the given conditions.
According to the problem, $f(x+t)=\frac{(x+t+1)^{2}}{4} \leqslant x \Rightarrow(x+t+1)^{2} \leqslant 4 x$ $\Rightarrow x^{2}+2(t-1) x+(t+1)^{2} \leqslant 0$ holds for all $x \in [1, m]$.
$$
\begin{array}{l}
\text { Thus }\left\{\begin{array}{l}
1+2(t-1)+(t+1)^{2} \leqslant 0, \\
m^{2}+2(t-1) m+(t+1)^{2} \leqslant 0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
-4 \leqslant t \leqslant 0, \\
(m+t-1)^{2} \leqslant(t-1)^{2}-(t+1)^{2}=-4 t
\end{array}\right. \\
\Rightarrow m-|t-1| \leqslant|m+t-1| \leqslant 2 \sqrt{-t} \\
\Rightarrow m \leqslant|t-1|+2 \sqrt{-t} \leqslant 9 .
\end{array}
$$

Equality holds when $t=-4$. In this case, for $\forall x \in [1,9], f(x-4)-x=\frac{(x-3)^{2}}{4}-x$ $=\frac{1}{4}\left[(x-3)^{2}-4 x\right]=\frac{1}{4}\left(x^{2}-10 x+9\right)=\frac{1}{4}(x-1)(x-9) \leqslant 0$.
In summary, the largest $m=9$.