 Solution

The multiplicative algebraic group ${\mathbb{G}}_m$ has coordinate ring $\mathbb{C}[t,t^{-1}]$. An algebraic morphism $f:{\mathbb{G}}_m\to {\mathbb{G}}_m$ therefore sends the invertible coordinate on the target to a unit of this Laurent polynomial ring.

Every unit of $\mathbb{C}[t,t^{-1}]$ has the form $c{t}^i$, with $c\in {\mathbb{C}}^{\ast }$ and $i\in \mathbb{Z}$. To prove this, suppose Laurent polynomials $P,Q$ satisfy $PQ=1$. Write their least and greatest active exponents as $a\le b$ and $c\le d$. The least and greatest exponents in their product are $a+c$ and $b+d$, since their endpoint coefficients are nonzero. Both must be zero. Consequentlyso each factor has only one active exponent. This proves the asserted unit description.

Thus $f(t)=c{t}^i$. A group homomorphism sends the identity to the identity, so $f(1)=1$ and $c=1$. Conversely, for every integer $i$, including negative integers and zero, $t\mapsto t^i$ is regular on ${\mathbb{G}}_m$ and satisfies $(st{)}^i=s^i{t}^i$. We obtainThese are exactly the characters of an algebraic torus in the one-dimensional case. The algebraic-morphism requirement is essential to the argument: the map must come from the Laurent coordinate ring.