Weak and weak* topology coincide for a non-reflexive space that is isomorpic to its dual? There are Banach spaces which are isomorphic to their second dual but not reflexive (most famously, the James space).
Now let $X$ be such a space and $X'$ be its dual space and let $\phi:X\to X''$ be an isomorphism between $X$ and $X''$. Now consider the weak* topology on $X'$, i.e. the topology generated by the neighborhood base
$$\tilde B_r(y_1,...y_n) = \{x\ :\ |x(y_j)| < r,\ j=1,...,n\},$$
where the $y_j$'s are elements in $X$.
This neighborhood base is in one-to-one correspondence with the neighborhood base for the weak topology on $X'$ i.e. with the sets
$$B_r(f_1,…,f_n) = \{x\ :\ |f_j(x)| < r,\ j=1,…,n\},$$
where the $f_j$'s are elements in $X''$ (via the isomorphism $\phi$).
However, this seem to contradict the fact that the weak and the weak* topology coincide precisely when $X$ is reflexive!?
There should be some flaw in my reasoning but somehow I am blocked here and would be glad in somebody could help out…
 A: Let $\phi$ be an “unnatural” isometric isomorphism between $X$ and $X''$. If $X$ is not reflexive, then there exists some $y\in X$ such that for all $z\in X$, there is some $x_z\in X'$ for which $\phi(y)(x_z)\neq x_z(z)$. This destroys the possibility to identify $\tilde B_r(y)$ with $B_r(\phi(y))$ through $\phi$.
ADDED: Here is a somewhat more transparent proof that the two neighborhood bases cannot be identified through $\phi$. If you assume that the two neighborhood bases can be identified through $\phi$, then you have, in particular, that $\tilde B_r(w)=B_r(\phi(w))$ for all $w\in X$ and $r>0$. That is, for any $x\in X'$, $w\in X$, and $r>0$, $|x(w)|<r$ if and only if $|\phi(w)(x)|<r$. Therefore, $x(w)$ and $\phi(w)(x)$ must have the same absolute values for all $x\in X'$ and $w\in X$:
$$|\phi(w)(x)|=|x(w)|=|\psi(w)(x)|\quad\forall x\in X',w\in X,\tag{$*$}$$
where $\psi$ is the canonical embedding.
Now pick any $f\in X''$. Then, there exists unique $w\in X$ such that $f=\phi(w)$ since $\phi$ is an (“unnatural”) isometric isomorphism between $X$ and $X''$. By ($*$), $\phi(w)$ and $\psi(w)$ (considered as continuous linear functionals over $X^*$) have the same kernel, so they must be proportional (for an easy proof of this, see here): there exists some $c\in\mathbb F$ (where $\mathbb F$ denotes the field over which $X$ is a vector space) such that $\phi(w)=c\psi(w)=\psi(cw)$. Consequently, $f=\psi(cw)$. Since $f$ has been an arbitrary element of $X''$, it follows that $X''=\psi(X)$. In other words, the canonical embedding $\psi$ is surjective, so $X$ must be reflexive—a contradiction.
This contradiction reveals that there must exist some $x\in X'$, $w\in X$, and $r>0$ such that either $|x(w)|<r$ and $|\phi(w)(x)|\geq r$, or $|x(w)|\geq r$ and $|\phi(w)(x)|< r$.
