10. Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}$ and $a \neq 0)$ satisfy the following conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$;
(2) When $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0. Find the maximum value of $m(m>1)$ such that there exists $t \in \mathbf{R}$, for any $x \in[1, m]$, $f(x+t) \leqslant x$.
10. From $f(x-4)=f(2-x)$, we know: the axis of symmetry is $x=-1$. Therefore, $-\frac{b}{2a}=-1$, which means $b=2a$.

From (3), we know: when $x=-1$, $y=a-b+c=0$. From (1), we get: $f(1) \geqslant 1$, and from (2), we get: $f(1) \leqslant 1$. Therefore, $f(1)=1$, which means $a+b+c=1$. From this, we can derive: $\left\{\begin{array}{l}b=\frac{1}{2} \\ a=c=\frac{1}{4}\end{array}\right.$.

If 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 $f(t+m) \leqslant m$, which means $\frac{1}{4}(t+m)^{2}+\frac{1}{2}(t+m)+\frac{1}{4} \leqslant m$. Simplifying, we get: $m^{2}-2(1-t)m+(t^{2}+2t+1) \leqslant 0$. Solving this, we get: $1-t-\sqrt{-4t} \leqslant m \leqslant 1-t+\sqrt{-4t}$. Therefore, $m \leqslant 1-t+\sqrt{-4t} \leqslant 1-(-4)+\sqrt{16}=9$.

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.