Must a multiplicative bijection on a ring $R \rightarrow R'$ with all elements nilpotent or units be an isomorphism? $f : R \rightarrow R'$ is a multiplicative bijection, i.e. for all $x,y \in R$ we have $f(xy)=f(x)f(y)$, where $R$ is a ring such that every element is either a unit or nilpotent.
Must $f$ be an isomorphism?
Thoughts:
If this were to be true, it is sufficient to prove $f(x+y)=f(x)+f(y)$ for any $x,y$. So it would be nice to relate the addition and multiplication operations together somehow. My first thought to obviously do this is some bracket expansion of the form
$f((x+y)^k)=(f(x+y))^k$ and relate somehow. (We know that the sum of nilpotents is nilpotent by such a bracket expansion. This leads to the nilpotents form an ideal which might be useful?).
I do not see how to proceed with this idea though. We could attempt some downwards induction on 
$[f(x)+f(y)]^k = [f(x+y)]^k$
since we have a base case by nilpotence. But this seems unlikely as we do not know we are in an integral domain or inverses exist so cant cancel factors easily to reduce to $k-1$ case.
I have heard that there is a unique prime ideal in such a ring, but don't know how I can utilise this.
N.B. The original question specified every element was nilpotent. But then we must have $R$ is trivial ring so is clearly true. I would therefore guess they mean every non unit element? 
Any ideas/hints would be appreciated!
 A: Let $R=\mathbb Z_2[x,y]/\langle x^2,xy,y^2\rangle$. Its elements are
$$
  R=\{0,x,y,x+y,1,1+x,1+y,1+x+y\}.
$$
Let $N=\{0,x,y,x+y\}$. Then $N^2=0$ and $R=N\cup(1+N)$. In particular the elements of $N$ are nilpotent, and those of $1+N$ are units.
Let $f:R\to R$ be the permutation that swaps $y,y+x$ and fixes all other elements. Note that $f(N)=N$ and $f$ is the identity on $1+N$. Then
$$
  f(ab)=f(a)f(b)
$$
for all $a,b\in R$. Indeed if $a,b\in N$ then $ab=0$ and $f(a)f(b)=0$. If $a,b\in 1+N$ then $ab\in 1+N$ and $f(ab)=ab=f(a)f(b)$. Finally if $a\in N$ and $b\in 1+N$ then $ab=a$ and $f(a)f(b)=f(a)$.
However $f$ is not additive, since
$$
  f(1+y)=1+y\neq 1+x+y=f(1)+f(y).
$$
A: If you want an example with non-isomorphic rings take
$$\begin{align*}\mathbb F_2[x]/x^2 &\to \mathbb Z/4 \\ 0 &\mapsto 0 \\ 1 &\mapsto 1 \\ x &\mapsto 2 \\ 1 + x &\mapsto 3.\end{align*}$$
I leave it to you to see that this is multiplicative, each of these rings satisfies the condition that every element is either nilpotent or a unit, and these rings are not isomorphic.
