Given a string of length $2n$, we perform the following operation:
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        [*]Place all the even indexed positions together, and then all the odd indexed positions next. Indexing is done starting from $0$.[/*]
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    For example, say our string is ``abcdef''. Performing our operation yields ``abcdef'' $\to$ ``acebdf''. Performing the operation again yields ``acebdf'' $\to$ ``aedcbf''. Doing this repeatedly, we have:\\
    ``abcdef'' $\to$ ``acebdf'' $\to$ ``aedcbf'' $\to$ ``adbecf'' $\to$ ``abcdef''.\\\\
    You can assume that the characters in the string will be unique. It can be shown that, by performing the above operation a finite number of times we can get back our original string.\\\\
    Given $n$, you have to determine the minimum number of times the operation must be performed to get our original string of length $2n$ back.\\\\
    In the example given above, $2n = 6$. The minimum steps required is $4$.
To determine the minimum number of times the operation must be performed to get our original string of length \(2n\) back, we need to analyze the transformation in detail.

1. **Define the Operation:**
   The operation involves placing all characters at even indices first, followed by characters at odd indices. For a string \(S\) of length \(2n\), the transformation can be described as:
   \[
   S = s_0 s_1 s_2 \ldots s_{2n-1} \rightarrow s_0 s_2 s_4 \ldots s_{2n-2} s_1 s_3 s_5 \ldots s_{2n-1}
   \]

2. **Reverse Operation:**
   The reverse operation \(f\) maps the character at index \(j\) to a new index \(f(j)\). Specifically, the character at index \(j\) moves to:
   \[
   f(j) \equiv 2j \mod (2n-1)
   \]
   This is because the operation effectively doubles the index and wraps around modulo \(2n-1\).

3. **Finding the Minimum \(k\):**
   We need to find the smallest \(k\) such that applying the operation \(k\) times returns the string to its original order. Mathematically, this means finding \(k\) such that:
   \[
   f^k(j) = j \quad \text{for all } j
   \]
   This implies:
   \[
   2^k j \equiv j \mod (2n-1)
   \]
   Simplifying, we get:
   \[
   (2^k - 1)j \equiv 0 \mod (2n-1)
   \]
   Since this must hold for all \(j\), the smallest \(k\) is the order of 2 modulo \(2n-1\), denoted as \(\text{ord}_2(2n-1)\).

4. **Conclusion:**
   The minimum number of times the operation must be performed to return the string to its original order is \(\text{ord}_2(2n-1)\).

The final answer is \( \boxed{ \text{ord}_2(2n-1) } \).