Morphisms in the category of rings We know that in the category of (unitary) rings, $\mathbb{Z}$ is the itinial object, i.e. it is the only ring such that for each ring $A,$ there exists a unique ring homomorphism $f:\mathbb{Z} \to A$.
This means, in particular, that $\mathbb{R}$ does not satisfy this property, so for a certain ring $B$, we can construct $g:\mathbb{R} \to B$ and $h:\mathbb{R} \to B$ such that $g \ne h$.
So far, I have proven that $B$ cannot be an ordered ring and cannot be $\mathbb{R}^n \ (n\in\mathbb{N})$. Can you help me finding such a ring $B$?
 A: Many useful and (sometimes surprisingly) true things have been said in other answers; let me just make some simple remarks.
To begin, you could have consulted the proof of what you implicitly used, namely that if $\def\Z{\Bbb Z}\Z$ is an initial object, then it is automatically unique, up to canonical isomorphism (indeed, for once, up to unique isomorphism). The (very standard) proof goes: suppose $R$ is another initial object in the category, then by assumption there are unique morphisms $f:\Z\to R$ and $g:R\to\Z$. Then $g\circ f:\Z\to\Z$ is a morphism and $\Z$ being initial it is unique and therefore equal to the identity of$~\Z$; similarly $f\circ g:R\to R$ is a morphism and $R$ being initial it is unique and therefore equal to the identity of$~R$, from which it follows that $f$ and $g$ are inverse morphisms, so $R$ is isomorphic to $Z$.
Now taking $\def\R{\Bbb R}R=\R$ it is clear that $f:\Z\to\R$ fails to be surjective so cannot have an inverse; by (the contrapositive of) the above argument this shows that $\R$ cannot be initial, and it does not admit a morphism $g:\R\to\Z$ at all. [That is a small lie. The argument uses the hypothesis that $R$ is initial twice: for the existence of $g$, and for equating $f\circ g$ to the identity of$~R$; one or the other must fail. But it is fairly easy to see that for $R=\R$ the former fails, and indeed one does have the property that any morphism $\R\to\R$ equals the identity, though it takes a bit of works to show that.]

Next, and rather unrelated, some words about the tensor product proof that $\R$ does not even have, as the above might suggest, the weaker property that if a morphism $\R\to A$ exists, then it is unique. You can think of the $\otimes$ in the construction of the ring $S\otimes S$ as similar to the fraction bar used to construct $\Bbb Q$, with the following difference: in fractions we decree that one can cancel (or introduce) equal factors ($an/nb=a/b$), but for the tensor product one decrees that one can move a factor across the $\otimes$ symbol in either direction ($an\otimes b=a\otimes nb$). The difference has as consequence that while sums of fractions can always be combined to a single fraction, sum of tensors cannot always.
The factors allowed to be moved are integers (the bare minimum), or possibly some larger class $R$ of scalars, which must then be attached as $\otimes_R$ to indicate this attribute of the construction; it serves as a gatekeeper that allows elements of (the ring) $R$ to cross the $\otimes_R$, but not others. Nonetheless, even for a bare $\otimes_\Z$, one can effectively pass rational factors across: to move $p/q$ from left to right, it suffices to move a factor $p$ from left to right and a factor $q$ from right to left (therefore $\otimes_\Bbb Q$ is not really different from $\otimes_\Z$). However one cannot go beyond rationals: $\sqrt2\otimes_\Z1$ and $1\otimes_\Z\sqrt2$ are different, for if they could be shown equal using a finite number of tensor-equivalences, one would basically have produced a fraction $p/q$ equal to $\sqrt2$, which is impossible. (A formal proof requires work.)
As a concluding remark: the ring $\Bbb Q$ and the finite fields $\Z/p\Z$ with $p$ prime do have the property that morphisms from them are unique if they exist. These are exactly the prime fields: fields without any proper sub-fields.
A: It is easier to show that for a certain ring $B$ there is no ring homomorphism at all $\mathbb R\to B$. In fact we can take $B=\mathbb Z$.
Assume to the contrary that $h:\mathbb R\to\mathbb Z$ is a homomorphism. We seek a contradiction.
If there is any nonzero $a\in\mathbb R$ such that $h(a)=0$, then we're done, because then $h(1)=h(a^{-1}a)=h(a^{-1})h(a)=h(a^{-1})\cdot 0 = 0$, which is not allowed for a unitary ring homomorphism.
Let's consider what $h(\sqrt 2)$ is. As argued above it can't be $0$, so it must be some nonzero integer $n$. Let's assume that $n$ is positive (the other case goes similarly). Then
$$ \underbrace{h(\sqrt2/n)+\cdots+h(\sqrt2/n)}_{n\text{ times}} = h(\underbrace{\sqrt2/n+\cdots+\sqrt2/n}_{n\text{ times}}) = h(\sqrt2)=n$$
but the only way this can be true is if $h(\sqrt2/n)=1$, and therefore
$$ h(\sqrt2/n-1) = h(\sqrt2/n)-h(1) = 1-1 = 0 $$
However, $a=\sqrt2/n-1$ is irrational and therefore in particular not zero, so this leads to a contradiction as above.
A: In fact, there is a universal ring with two distinct morphisms from $\mathbb{R}$, the ring $\mathbb{R}\otimes_\mathbb{Z}\mathbb{R}$.
I am unsure of whether it has nicely-presented quotients with the same property.
A: $\mathbb{R}$ has two distinct embeddings into the tensor product $\mathbb{R}\otimes_{\mathbb{Z}} \mathbb{R}$, given by $r\mapsto r\otimes 1$ and $r\mapsto 1\otimes r$. 
These are the two inclusions coming from the fact that the tensor product is the coproduct in the category of rings.
A: Let $\phi:\Bbb R\to B$ be a ring morphism. Then $\ker\phi$ is an ideal of $\Bbb R$. But the only ideals of $\Bbb R$ are $\Bbb R$ and $\langle 0\rangle$. Thus $\ker\phi=\Bbb R$ or $\ker\phi=0$.
Suppose $\ker\phi=\Bbb R$. Then $B$ must be the trivial ring since $\phi(1)=1_B$ and the only ring satisfying $1_B=0_B$ is the trivial ring. In this case $\phi:\Bbb R\to B$ is unique.
Suppose $\ker\phi=0$. Then $\phi$ is injective. It follows that $\DeclareMathOperator{image}{image}\image(\phi)$ is a subring of $B$ isomorphic to $\Bbb R$. 
This proves that every nontrivial ring $B$ without a subring isomorphic to $\Bbb R$ admits no homomorphisms $\phi:\Bbb R\to B$.
Examples include $B=\Bbb Z$, $B=\Bbb Z_n$, $B=\Bbb Q$.
A: Here's an answer to the original question which doesn't use tensor products (but does use some field theory).
Let $i$ be the usual inclusion of $\mathbb{R}$ into $\mathbb{C}$, and let $\sigma$ be any automorphism of $\mathbb{C}$ which does not fix $\mathbb{R}$ pointwise. Then $\sigma\circ i\neq i$ are distinct ring homomorphisms $\mathbb{R}\to\mathbb{C}$.
To find an automorphism of $\mathbb{C}$ which does not fix $\mathbb{R}$ pointwise, you could, for example,


*

*Extend the automorphism of $\mathbb{Q}[\sqrt{2}]$ given by $\sqrt{2}\mapsto -\sqrt{2}$ to an automorphism of $\mathbb{C}$.

*Pick a transcendence basis $\mathcal{B}$ of $\mathbb{C}$ over the algebraic closure of $\mathbb{Q}$ which contains $\pi$. Then any permutation of $\mathcal{B}$ which moves $\pi$ extends to an automorphism of $\mathbb{C}$.

A: You're wrong in negating the statement of being an initial object.
A ring $R$ is an initial object if and only if, for every ring $S$ there exists a ring homomorphism $R\to S$ and, if $f,g\colon R\to S$ are ring homomorphisms, then $f=g$.
Negating this becomes

$R$ is not an initial object if and only if there exists a ring $S$ such that there is no ring homomorphism $R\to S$ or there are two distinct homomorphisms $R\to S$.

For instance, the ring $\mathbb{F}_2$ has the property that, if a ring homomorphism $f\colon\mathbb{F}_2\to S$ exists, then it is unique. The reason is obvious: $f(0)=0$ and $f(1)=1$, plus the fact that $\mathbb{F}_2=\{0,1\}$.
So there is no ring $S$ with two distinct ring homomorphisms $f,g\colon\mathbb{F}_2\to S$, but $\mathbb{F}_2$ is not an initial object; indeed, there is no ring homomorphism $\mathbb{F}_2\to\mathbb{Z}$.
