Generalised Symmetric group and Semidirect Product

Question

$G= \mathbb{Z}_r \wr \mathfrak{S}_n:=(\mathbb{Z}_r)^n\rtimes\mathfrak{S}_n$ is the generalised symmetric group where the element $G$ is denoted by $(f,\pi)$ where $f:\{1,2,\dots n\}\to \mathbb{Z}_r$ (ie., an element of $(\mathbb{Z}_r)^n)$ and $\pi\in\mathfrak{S}_n$. The group operation $$(g,\sigma)\cdot(f,\pi)=(g\cdot (f\circ\sigma^{-1}), \sigma\pi).$$ I understand that this is the usual semidirect product operation.

We can think symmetric group as two ways in which it permutes $n$ boxes (corresponds to natural left-right actions of $\mathfrak{S}_n$ ). Let us say $n=4$ and we have boxes numbers(value) 1,2,3,4 placed in for corner vertices of square named (position) I,II,III,IV. Then $\mathfrak{S}_4$ acts either by

1. by permuting positions I,II,III,IV (this corresponds to natural left action of $\mathfrak{S}_4$)
2. or by permuting values or box numbers 1,2,3,4. (this corresponds to natural right action of $\mathfrak{S}_4$)

I belive just like we think symmetric group by its action of permuting $n$ boxes. The generalised symmetric group can also be thought as "sort of permuting n-groups" instead of boxes.

Lets us say $r=3$ now one may think we have stuck equilateral triangle in the corner vertices of square. See the attached figure. Now we can rotate any of these triangle and also permute these triangles. A symmetry of this object is pair of operation

1. Rotating each triangles
2. and permuting triangles.

This is exactly encoded in the data $(f,\pi)\in G$. But my worry is how did we encoded this data?

• Qn 01: what do we mean by $f(i)$? is $f(i)$ an operation to performed on triangle in the $I^{th}$ position or the operation to be performed to $ i^{th} $ triangle?
• Qn 02: How does $\mathfrak{S}_4$ acts? whether by permuting positions or the value/numbers.
• Qn 03: How do we see that operation in the semi direct product is natural in this context? that is by First rotate triangles by $f(i)$ then permute by $\pi$ and then again rotate by $g(j)$ and finally permute $\sigma$ by is the same as the product given in semidirect product.

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PS: These are so confusing and had spent a lot of time to fix this before bringing this here. Kindly help me with this. Thanks!