Since $f(x-4) = f(2-x)$ for $x \in \mathbb{R}$, it is known that the quadratic function $f(x)$ has $x = -1$ as its axis of symmetry. By condition (3), we know that $f(x)$ opens upward, that is, $a > 0$. Hence
$$f(x) = a(x+1)^2 \quad (a > 0).$$
By condition (1), we get $f(1) \ge 1$ and by (2), $f(1) \le \left(\frac{1+1}{2}\right)^2 = 1$. It follows that $f(1) = 1$, i.e. $a(1+1)^2 = 1$. So $a = \frac{1}{4}$.
Thereby, $f(x) = \frac{1}{4}(x+1)^2$.

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Since the graph of the parabola $f(x) = \frac{1}{4}(x+1)^2$ opens upward, and a graph of $y = f(x+t)$ can be obtained by translating that of $f(x)$ by $t$ units. If we want the graph of $y = f(x+t)$ to lie under the graph of $y = x$ when $x \in [1, m]$, and $m$ to be maximal, then $1$ and $m$ should be two roots of an equation with respect to $x$.
$$\frac{1}{4}(x+t+1)^2 = x. \qquad \textcircled{1}$$
Substituting $x = 1$ into (1), we get $t = 0$ or $t = -4$.
When $t = 0$, substituting it into (1), we get $x_1 = x_2 = 1$ (in contradiction with $m > 1$).
When $t = -4$, substituting it into (1), we get $x_1 = 1$, and $x_2 = 9$; and so $m = 9$.
Moreover, when $t = -4$, for any $x \in [1, 9]$, we have always
$$(x-1)(x-9) \le 0 \\ \Leftrightarrow \frac{1}{4}(x-4+1)^2 \le x,$$
that is
$$f(x-4) \le x.$$
Therefore, the maximum value of $m$ is $9$.