Can you have a ring homomorphism from a ring to itself which isn't the identity? By a ring I mean a ring with a multiplicative identity. To me, at this point, this sounds like a fairly simple question, but I haven't been able to come up with any such homomorphism, nor has searching Google for one proved fruitful.
 A: I post my comment as suggested by Davide. 
Take $R=\mathbb C$ and the complex conjugation as ring automorphism.
A: Sure, let $R$ be a commutative ring with identity and consider the map $R[x] \to R[x]$ determined by $p(x) \mapsto p(0)$.
To see a case where the map is an isomorphism, let $R = \Bbb Z[\sqrt{2}]$ and consider the map $a + b \sqrt{2} \mapsto a - b \sqrt{2}$. You should check that this is a homomorphism, and actually gives an isomorphism from $\Bbb Z[\sqrt{2}]$ to itself.
A: You can consider for example $M_n(\mathbb{R}),$ the ring of $n\times n$ matrices over the real numbers (with $n>1$) and take the map $M_n(\mathbb{R})\ni A \mapsto PAP^{-1}\in M_n(\mathbb{R}),$ where $P\ne xI$ is an invertible matrix of $M_n(\mathbb{R})$ and $x\in \mathbb{R}.$
Of course you can take any ring $R$ instead of $\mathbb{R}.$
A: Recall that functors between categories preserve automorphisms. So for any faithful functor $F : \mathbf{Set} \rightarrow \mathbf{C}$, the object $F(\kappa)$ will have at least $\kappa!$-many distinct automorphisms, where I write $\kappa!$ for the number of set-theoretic automorphisms of a set with $\kappa$-many elements. For example, let $F : \mathbf{Set} \rightarrow \mathbf{CRing}$ denote the free functor. Then $F(2)$ has at least $2! 
\;(=2)$ different automorphisms (one for each permutation of the variables!), and $F(\aleph_0)$ has at least $\aleph_0!$-many (that is, $2^{\aleph_0}$-many; that is, continuum-many) distinct automorphisms.
