First, let's show that $$\sum_{k=0}^{n}\left(\left(1-\frac{2k}{n}\right)\binom{n}{k}\right)^2=2\sum_{k=0}^{\lfloor{\frac{n-1}{2}}\rfloor}\left(\left(1-\frac{2k}{n}\right)\binom{n}{k}\right)^2$$

If $n$ is even, let $n=2m$. Then $\lfloor{\frac{n-1}{2}}\rfloor=m-1$. Then $$\begin{align*} &\sum_{k=m}^{n=2m}\left(\left(1-\frac{2k}{n}\right)\binom{n}{k}\right)^2 \\=\ &\sum_{k=m+1}^{n=2m}\left(\left(1-\frac{2k}{n}\right)\binom{n}{k}\right)^2 + \left(\left(1-\frac{2m}{n}\right)\binom{n}{k}\right)^2 \\ =\ &\sum_{k=m+1}^{n=2m}\left(\left(1-\frac{2k}{n}\right)\binom{n}{k}\right)^2 \\=\ &\sum_{k=0}^{m-1}\left(\left(1-\frac{2(n-k)}{n}\right)\binom{n}{n-k}\right)^2 \\=\ &\sum_{k=0}^{m-1}\left(\left(\frac{2k}{n}-1\right)\binom{n}{n-k}\right)^2 \\=\ &\sum_{k=0}^{m-1}\left(\left(1-\frac{2k}{n}\right)\binom{n}{k}\right)^2\end{align*}$$

This means that the sum of first $m$ terms (when $k$ enumerates from $0$ to $m-1$) equals the sum of the last $m$ terms (when $k$ enumerates from $m$ to $2m$), and the middle term equals $0$ (i.e. $k=m$). Therefore, the proposed identity holds.

When $n$ is odd, let $n=2m+1$. Then $\lfloor{\frac{n-1}{2}}\rfloor=m$. Using the same transformation can show that the sum of the first $m$ terms equals the sum of the next $m$ terms. In this case, there is no middle term. The proposed identity will still hold.

Therefore, it is sufficient to show the following result in order to prove the original claim $$\sum_{k=0}^{n}\left(\left(1-\frac{2k}{n}\right)\binom{n}{k}\right)^2=\frac{2}{n}\binom{2n-2}{n-1}$$

Let's expand this expression into three terms $$\sum_{k=0}^{n}\left(\left(1-\frac{2k}{n}\right)\binom{n}{k}\right)^2=\underbrace{\sum_{k=0}^{n}\binom{n}{k}^2}_{A}-\underbrace{\frac{4}{n}\sum_{k=0}^{n}k\binom{n}{k}^2}_{B} +\underbrace{\frac{4}{n^2}\sum_{k=0}^{n}k^2\binom{n}{k}^2}_{C}$$

Utilizing the conclusion of # 3976 (or employing the Vandermonde's identity directly) yields $$A=\binom{2n}{n}$$

The value of $B$ can also be computed  using the Vandermonde's identity: $$B=\frac{4}{n}\sum_{k=0}^{n}k\binom{n}{k}^2 = {4}\sum_{k=1}^{n}\binom{n-1}{k-1}\binom{n}{k} ={4}\sum_{k=1}^{n}\binom{n-1}{k-1}\binom{n}{n-k} = 4\binom{2n-1}{n-1}$$

The value of $C$ can be calculated in a similar way: $$\frac{4}{n^2}\sum_{k=0}^{n}k^2\binom{n}{k}^2=4\sum_{k=1}^n\binom{n-1}{k-1}^2=4\binom{2n-2}{n-1}$$

It follows that $$\begin{align*} A-B+C =\ &\binom{2n}{n} - 4\binom{2n-1}{n-1} + 4\binom{2n-2}{n-1} \\=\ &\binom{2n}{n}-4\binom{2n-2}{n-2}\\=\ &\frac{(2n)!}{n!n!} - 4\cdot\frac{(2n-2)!}{(n-2)!n!} \\=\ &\frac{(2n-2)!}{(n-2)!n!}\left(\frac{(2n)(2n-1)}{n(n-1)}-4\right)\\=\ &\frac{2(2n-2)!}{n!(n-1)!} \\=\ &\frac{2}{n}\binom{2n-2}{n-1} \end{align*}$$