We know every Hilbert Space $\mathcal{H}$ is isometrically isomorphic to its dual, not linearly but antilinearly. This is carried out by defining $\mathcal{H}_*$ to be same $\mathcal{H}$ in the sense of additive group and by altering the scalar multiplication in $\mathcal{H}$. We define $\lambda.x=\bar{\lambda}x$ in $\mathcal{H}_*$ and a new inner product $\langle x,y \rangle_*=\langle y,x \rangle$. Then $\mathcal{H}_*$ is a Hilbert space with the norm being the same as $\mathcal{H}$. Then the map $v:\mathcal{H}\rightarrow (\mathcal{H}_*)^*$ such that $v(x)y=\langle y,x \rangle_*=\langle x,y \rangle$. This map is the desired isometric isomorphism.
Now my question is that if this property is satisfied by any Banach space $X$, then can it be made into a Hilbert space?
I would just like to mention that the examples I could think of were mainly $\mathcal{L}_2(X,\mu)$ and the space of Hilbert Schmidt operators $\mathcal{L}_2\mathcal{(H)}$.
