Solution Set for Composite Function Equation Involving Sine and Square Functions

To find the set of all $x \in \mathbb{R}$ satisfying the equation
$(f \circ g \circ g \circ f)(x) = (g \circ g \circ f)(x)$
where $f(x) = x^2$ and $g(x) = \sin x$, we proceed as follows:

Let $u = (g \circ g \circ f)(x)$.

Then the given equation can be written as:
$f((g \circ g \circ f)(x)) = (g \circ g \circ f)(x)$
$f(u) = u$

Since $f(x) = x^2$, this gives:
$u^2 = u \implies u(u - 1) = 0$
$\implies u = 0 \quad \text{or} \quad u = 1$

Now, recall that:
$u = g(g(f(x))) = \sin(\sin(x^2))$

Case 1: $u = 0$
$\sin(\sin(x^2)) = 0$
The sine function vanishes at integer multiples of $\pi$, so:
$\sin(x^2) = m\pi, \quad m \in \mathbb{Z}$
Since the range of the sine function is $[-1, 1]$, and the only integer multiple of $\pi$ in this interval is $0$ (as $\pi \approx 3.14159 > 1$), we must have:
$m = 0 \implies \sin(x^2) = 0$
This implies:
$x^2 = n\pi, \quad n \in \mathbb{Z}$
Since $x \in \mathbb{R}$, we have $x^2 \ge 0$, which restricts $n$ to non-negative integers:
$n \in \{0, 1, 2, \dots\}$
Thus:
$x = \pm\sqrt{n\pi}, \quad n \in \{0, 1, 2, \dots\}$

Case 2: $u = 1$
$\sin(\sin(x^2)) = 1$
The sine function equals $1$ when:
$\sin(x^2) = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}$
However, $\frac{\pi}{2} \approx 1.5708 > 1$, and for any other integer $k$, the magnitude of $\frac{\pi}{2} + 2k\pi$ is strictly greater than $1$. Since $-1 \le \sin(x^2) \le 1$ for all real $x$, there are no real solutions for this case.

Combining the results, the set of all real solutions is:
$x = \pm\sqrt{n\pi}, \quad n \in \{0, 1, 2, \dots\}$

Hence, the correct option is A.