Complex hilbert space and its complex conjugate vector space. Let $V$ be a complex Hilbert space, and $\overline{V}$ its conjugate vector space (see https://en.wikipedia.org/wiki/Complex_conjugate_vector_space)
What are sufficient conditions such that $V \cong \overline{V}$. Is it always true? If not, can you provide a counterexample?
On the wikipedia page, it is written that $V$ and $\overline{V}$ are isomorphic vector spaces, because they have the same dimension. It seems that this argument only works when the spaces are finite dimensional.
 A: Yes, they are isomorphic.  But not canonically isomorphic.  
Again, this is because they have the same "dimension", where now "dimension" is the "Hilbert dimension": the cardinality of a maximal orthonormal set.
A: Let the map $\varphi:\mathcal{H}\to\overline{\mathcal{H}}$ from a Hilbert space $\mathcal{H}$ into its complex conjugate $\overline{\mathcal{H}}$ be antilinear. Then 
$$\varphi(x+y)=\varphi(x)+\varphi(y)\hspace{0.2cm}\text{and}\hspace{0.2cm}\varphi(\alpha x)=\overline{\alpha}\varphi(x)$$
for all $x,y\in\mathcal{H}$. This map is also a homomorphism between $(\mathcal{H},\cdot)$ and $(\overline{\mathcal{H}},*)$ where $\alpha *x=\overline{\alpha}\cdot x$. Next define the map $\psi:\overline{\mathcal{H}}\to\mathcal{H}$ by $\psi(x):=\varphi^{-1}(x)$ for all $x\in\overline{\mathcal{H}}$. Then $\psi$ is an antilinear map as well since
$$\psi(x+y)=\varphi^{-1}(x+y)=\varphi^{-1}(\varphi(x')+\varphi(y'))=\varphi^{-1}(\varphi(x'+y'))=x'+y'$$ 
where we used the antilinearity of $\varphi$. But
$$x'+y'=\varphi^{-1}\varphi(x')+\varphi^{-1}\varphi(y')=\varphi^{-1}(x)+\varphi^{-1}(y)=\psi(x)+\psi(y)$$
hence $$\psi(x+y)=\psi(x)+\psi(y)$$
Similarly we have for the second property
$$\psi(\alpha*x)=\varphi^{-1}(\alpha*x)=\varphi^{-1}(\overline{\alpha}\cdot x)=\varphi^{-1}(\varphi(\alpha\cdot x))=\alpha\cdot x$$
thus $\psi$ is as well a homomorphism. Therefore any antilinear map $\varphi$ is a isomorphism between the algebraic structures $(\mathcal{H},\cdot)$ and  $(\overline{\mathcal{H}},*)$. So they must be isomorphic as vector spaces as well since vector spaces are algebraic structures themselves. 
