Let $k$ be an algebraically closed field and let $\mathfrak m$ be a maximal ideal in $R=k[x_1,\dots,x_r]$. Show that there is a $k$-algebra automorphism of $R$ taking $\mathfrak m$ to $(x_1,\dots,x_r)$.
Since $k$ is algebraically closed, every maximal ideal $\mathfrak m$ is of the form $(x_1-a_1,\dots,x_n-a_n)$. So would the map be defined as follows? Let $\phi: R\to R$ be defined on $x_i$ as $x_i\mapsto x_i+a_i$ and let $\phi(a_nx^n+\dots a_1x+a_0)=a_n\phi(x_1)^n+\dots+a_1\phi(x)+a_0$. Then by construction, $\phi$ is a $k$-algebra automorphism. Does this look OK?