Automorphisms of ideals of $\mathbb{C}[t]$ Let $f(t)\in\mathbb{C}[t]$, and let $I_f$ be the ideal in $\mathbb{C}[t]$ generated by $f(t)$.
The ideal $I_f$ has a natural $\mathbb{C}$-algebra structure. My question is the following:

For which polynomials $f\in \mathbb{C}[t]$ is the group of ($\mathbb{C}$-algebra) automorphisms of $I_f$ trivial?

Some examples: 
$1)$ If $f(t)=t$, there are (of course) infinitely many automorphims, defined by $t\mapsto \alpha t$, for any $\alpha\in \mathbb{C}[t]-\{0\}$. This case is very simple to understand, since the ideal generated by $t$ is a free $\mathbb{C}$-algebra (without unit) and any homomorphism can be defined just by defining the image of $t$.
$2)$ If $f(t)=t(t-1)$ we were able to find only one automorphism, given by $f\mapsto f$, and $tf\mapsto f-tf$. It is an automorphims of order $2$.
One can observe that  it is enough to define the above in $f$ and $tf$, since these two elements generate $I_f$, as a $\mathbb{C}$-algebra (although the algebra generated by these two polynomials are not free, so, one needs to be careful in defining homomorphisms)
$3)$ As a last example, we were not able to find any non-trivial automorphism of the ideal generated by $f(t)=t(t-1)(t-2)$.
 A: We may assume that $f$ is not a unit.  Let $A_f=I_f+\mathbb{C}\cdot 1$, the free unital algebra on $I_f$; this embeds naturally in $\mathbb{C}[t]$.  Furthermore, $A_f$ and $\mathbb{C}[t]$ have the same fraction field, and $\mathbb{C}[t]$ is the integral closure of $A_f$ in its fraction field.  Thus every automorphism of $A_f$ extends uniquely to an automorphism of $\mathbb{C}[t]$.  It is now easy to see that automorphisms of $I_f$ are in bijection with automorphisms of $\mathbb{A}^1$ that map the closed subscheme defined by $f$ to itself.  Concretely, such an automorphism is a linear polynomial that permutes the roots of $f$ (counting multiplicity).  I don't know of a concise characterization of when any such automorphism must be trivial, but I expect this description of the automorphisms is good enough for most purposes.  In particular, for instance, if $f$ has more than one distinct root it is easy to show from this description that every automorphism of $I_f$ is given by multiplication by a root of unity conjugated by translation by the average of the roots.
