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:



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> For which polynomials $f\in \mathbb{C}[t]$ is the group of ($\mathbb{C}$-algebra) automorphisms of $I\_f$ trivial?
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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 $f(t)=t(t-1)(t-2)$.



Automorphisms of ideals of $\mathbb{C}[t]$
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.


