Edit: This answer gives a construction of $W$ which shows that it need not contain more than ${n+1\choose 2}$ elements (if $X$ is real), but does not answer the question posed, as Yuval's argument does.
The easiest way to construct a finite set $W\subset X$ which detects orthogonality is as follows:
Let $\{\alpha_1,\dots,\alpha_n\}$ be a basis for $X$. Let
$W=\{\alpha_j+c\alpha_k:j\leq k;c=0,1\text{ or }i\}.$
Then $|W|=n^2$. Suppose $T$ is a linear operator on $X$ such that $\|T\alpha\|=\|\alpha\|$ for every $\alpha\in W$. If $j\leq k$, then
$\langle T\alpha_j,T\alpha_k\rangle=\frac{1}{2}\left(\|T\alpha_j+T\alpha_k\|^2+i\|T\alpha_j+iT\alpha_k\|^2\right)-\left(\|T\alpha_j\|^2+\|T\alpha_k\|^2\right)$
$=\frac{1}{2}\left(\|T(\alpha_j+\alpha_k)\|^2+i\|T(\alpha_j+i\alpha_k)\|^2\right)-\left(\|T\alpha_j\|^2+\|T\alpha_k\|^2\right)$
$=\frac{1}{2}\left(\|\alpha_j+\alpha_k\|^2+i\|\alpha_j+i\alpha_k\|^2\right)-\left(\|\alpha_j\|^2+\|\alpha_k\|^2\right)$
$=\langle \alpha_j,\alpha_k\rangle.$
If $j>k$, the above calculation shows that $\langle T\alpha_k,T\alpha_j\rangle=\langle \alpha_k,\alpha_j\rangle$, hence $\langle T\alpha_j,T\alpha_k\rangle=\overline{\langle T\alpha_k,T\alpha_j\rangle}=\overline{\langle \alpha_k,\alpha_j\rangle}=\langle\alpha_j,\alpha_k\rangle.$ Thus $T$ preserves inner products, and $T$ is unitary. Notice that if $X$ is real, $W$ need not contain more than ${n+1\choose2}$ elements; simply choose the set above, without the $n\choose2$ elements $\alpha_j+i\alpha_k$, $j\leq k$.