Suppose we are given $s_1\geq s_2\geq \ldots \geq s_n>0$ and let $Q\in GL_n(\mathbb{C})$ satisfying the two conditions. Then 

$Q\overline{Q}=\lambda I_n$ 

for some $\lambda\in\mathbb{C}$ and hence, by transposing, $Q^*Q^T=\lambda I_n$. Pick $v_k\in\mathbb{C}^n$ such that 

$Q^*Qv_k = s_kv_k$

Then, $Q^*Q^T(Q^{-T}Qv_k) = Q^*Qv_k = s_kv_k = \lambda(Q^{-T}Qv_k)$ that is 

$\lambda Qv_k=s_k Q^Tv_k$

multiplying by $\overline{Q}$ on both sides and conjugate we obtain,

$\overline{\lambda} Q\overline{Q}\overline{v}_k = s_k Q Q^*\overline{v}_k$

Since, $Q\overline{Q}=\lambda I_n$ and $s_k\neq0$ we have, 

$Q Q^*\overline{v}_k = (|\lambda|^2/s_k)\overline{v}_k$

Moreover, $Q^*Q$ and $QQ^*$ have the same eigenvalues and the monotonicity conditions on $s_1,\ldots s_n$ ensure we have, 

$\frac{|\lambda|^2}{s_n}=s_1,\text{ } \frac{|\lambda|^2}{s_{n-1}}=s_2,\text{ }\ldots , \text{ } \frac{|\lambda|^2}{s_1}=s_n$

This shows the choice $s_n=s_{n-1}$, $s_1\neq s_2$ allows no such $Q$.