If $D$ is an integral domain, then show that every automorphism $f$ of $D[X]$ which is identity on $D$ is of the form $f(X)=cX+d$, where $c$ is a unit of $D$.
It is easy to show that a function of this form is a ring automorphism and is uniquely determined by the conditions. But why each automorphism is of this form? This is Exercise 14 on page 167 in Hungerford's Algebra.
Let $g = \sum_{k=0}^N b_k X^k$ be the polynomial for which $f(g) = X,$ where $b_N \ne 0.$ Then $$\sum_{k=0}^M b_k f(X)^k = X.$$ Since $D$ is an integral domain, $b_N$ will not cancel with any of the nonzero coefficients of $f$, so you will not be able to solve this equation unless $f(X)$ (and $g$) both have degree $1$.