Given an IFS \(\mathcal{F} = \{X,f_1,f_2,\ldots, f_n\}\), and a set \(S\), it is relatively simple to find a family of IFS's that are all Hutchinson equivalent to \(\mathcal{F}\) over \(S\).
Let \(G\) be a set of functions from \(X\) to \(X\) such that \(g(S) = S\) for every \(g \in G\).
Let \(g_1,g_2,\ldots,g_n \in G\), and define a new IFS \(\overline{\mathcal{F}} = \{X,f_1g_1,f_2g_2,\ldots, f_ng_n\}\).
We want to show that \(\mathcal{F}\) and \(\overline{\mathcal{F}}\) are Hutchinson equivalent.
Let \(H(E) = \bigcup_{i=1}^{n}{f_i(E)}\) be the Hutchinson operator for \(\mathcal{F}\) and let \(\overline{H}(E) = \bigcup_{i=1}^{n}{f_ig_i(E)}\) be the Hutchinson operator for \(\overline{\mathcal{F}}\).
Then \(\overline{H}(S) = \bigcup_{i=1}^{n}{f_i(g_i(S))} = \bigcup_{i=1}^{n}{f_i(S)} = H(S)\).
So \(\mathcal{F}\) and \(\overline{\mathcal{F}}\) are Hutchinson equivalent over \(S\).
The set of all \(\overline{\mathcal{F}} = \{X,f_1g_1,f_2g_2,\ldots, f_ng_n\}\) such that \(g_1,g_2,\ldots,g_n \in G\) will be called the Hutchinson equivalent family of \(\mathcal{F}\) generated by \(G\) over \(S\).
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