Isomorphic rings? Let $f:R\to R$ be a ring isomorphism and $M$ is an $R-$module. Define $*:R\times M\to M$ maps $(r,m)$ to $r*m=f(r)m$. Then with this scalar product, $M$ is a $R-$module, namely $M^f$. My question is, are $M$ and $M^f$ always isomorphic, I can't prove it.
 A: Here is a very simple example:
$\mathbb Z$ is a $\mathbb Z[x]$-module via $f \cdot c := f(0)c$
Now consider the isomorphism $\phi: \mathbb Z[x] \to \mathbb Z[x], x \mapsto x+1$. 
There are only two isomorphism of abelian groups $\mathbb Z \to \mathbb Z$ (multiplication with $\pm 1$) and it is easy to see that both do not behave well with the two module structures.
Hence the two $\mathbb Z[x]$-modules are not isomorphic.
Some details:
Let $h: \mathbb Z \to \mathbb Z$ be multiplication with $\pm 1$. We have
$$h(x \cdot c) = h(0c)=h(0)=0,$$
but
$$x * h(c) = (x+1) \cdot h(c) = h(c)= \pm c.$$
A: No, they are not isomorphic in general.
Let $C_3$ be the cyclic group of order $3$ and let $x$ be a fixed generator for it.  This has a group automorphism exchanging $x$ and $x^{-1}$ which extends to a ring automorphism of the group algebra $\mathbb{C}[C_3]$. Let's take this to be our $f$.
Now for $M$ let's take the $1$ dimensional complex representation of $C_3$ where $x$ acts by $e^{2\pi i/3}$.  The twisted version $M^f$ is still a one dimensional representation, but this time it's the one where $x$ acts by $e^{4\pi i/3}$.  In particular these representations have different characters and are non-isomorphic.
