Update: The lead paragraph has been changed to reflect the solution to a related question.

I asked the question Is dimension given by the Klee trick ever sharp? and it has been answered in the affirmative using two embeddings of a $0$-dimensional metric spaces in $\mathbb{R}^1$.

In the interests of trying to better understand the Klee Trick, I thought I might ask a slightly more concrete question along those lines.

Given a metric space $K$ that embeds into $\mathbb{R}^n$, define a $Klee_n(K)$ as the minimum value for $m$ such that for any two embeddings $f_1, f_2 : K \rightarrow \mathbb{R}^n$ and any embeddings $g_1,g_2: \mathbb{R}^n \rightarrow \mathbb{R}^m$ there exists with a homeomorphism $h:\mathbb{R}^m \rightarrow \mathbb{R}^m$ such that $$h(g_1(f_1(K)))=g_2(f_2(K)).$$

(The Klee trick says $Klee_n(K) \leq 2n$.)

For example, $Klee_3(\mathbb{S}^1)=4$, since if $g_1, g_2$ correspond to embeddings of distinct (hyperbolic) knots then we see $Klee_3(\mathbb{S}^1)> 3$, but any two embeddings of $\mathbb{S}^1$ into $\mathbb{R}^4$ can be unknotted via an isotopy.

The other question asked if there was a metric space $K$ such that $Klee_n(K)=2n.$ However, for this question, is there a $K$ such that $Klee_3(K)=5$?