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EDIT: As I now tried to provide an answer to this problem, I would appreciate any proof verification of my reasoning. Thank you very much!

In Katz and Mazur's book (available here, page 109, page 60 in the pdf), it is stated as an obvious statement that any representable moduli problem is rigid. Actually, being new to these notions, I fail to see how one can show this. Let me sum up the situation.

We consider the category $(\textbf{Ell})$ of elliptic curves over a base (maps are cartesian squares of elliptic curves). Let $\mathcal{P}$ be a moduli problem for elliptic curves (that is a contravariant functor from $(\textbf{Ell})$ to $(\textbf{Set})$). Assume that $\mathcal{P}$ is representable: there exists an elliptic curve $\mathbb{E}/\mathcal{M}(\mathcal{P})$ together with a functorial isomorphism $$\mathcal{P}(E/S)\simeq\operatorname{Hom}_{(\textbf{Ell})}(E/S,\mathbb{E}/\mathcal{M}(\mathcal{P}))$$ for any object $E/S$ of $(\textbf{Ell})$.

The statement we need to prove is the following.

Let $E/S$ be an object of $(\textbf{Ell})$ given with an element $a\in \mathcal{P}(E/S)$. Let $\epsilon:E/S\rightarrow E/S$ be an automorphism. By functoriality, it induces a map $\mathcal{P}(E/S) \rightarrow \mathcal{P}(E/S)$. Assume that $a$ is a fixed point of this map. Then, $\epsilon$ is the identity.

This statement is said to be explicitly visible from the description of representable problems. Thus I feel like the proof should be quite straightforward, but I fail to see the point. Actually, I may not understand to what extent "to be representable" is a powerful statement in this context. Could somebody please clarify this point to me? I am thankful in advance for your help.

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I think I figured out a way to show it, however I am pretty unsure about this resolution. Any feedback would be gladly appreciated.

Assume that $\mathcal{P}$ is represented by a universal elliptic curve $\mathbb{E}/\mathcal{M}(\mathcal{P})$ together with a universal element $a_0\in \mathcal{P}(\mathbb{E}/\mathcal{M}(\mathcal{P}))$. This universal element is induced by the identity morphism on the universal elliptic curve.

Then, to give an element $a\in \mathcal{P}(E/S)$ is the same as to give a morphism $E/S\rightarrow \mathbb{E}/\mathcal{M}(\mathcal{P})$ in $(\textbf{Ell})$ such that the image of $a_0$ by the induced map $\mathcal{P}(\mathbb{E}/\mathcal{M}(\mathcal{P}))\rightarrow \mathcal{P}(E/S)$ is precisely $a$. That is the data of a morphism $\phi: S\rightarrow \mathcal{M}(\mathcal{P})$ and an isomorphism $\iota: E\xrightarrow{\sim} \mathbb{E}\times_{\mathcal{M}(\mathcal{P})}S$ of elliptic curves over $S$, such that the pullback of $a_0$ is $a$.

Now, $\operatorname{Aut(E/S)}$ acts on $\mathcal{P}(E/S)$ by keeping $\phi$ as it is and by composing $\iota$ on the right. To say that an automorphism $\epsilon$ has a fixed point means that $\iota\circ\epsilon=\iota$ as isomorphisms of elliptic curves $E\xrightarrow{\sim} \mathbb{E}\times_{\mathcal{M}(\mathcal{P})}S$ over $S$.

Thus, $\iota(\epsilon-1)=0$. Because $\iota$ is an isomorphism, $\operatorname{deg}(\iota)\not = 0$, so necessarily $\operatorname{deg}(\epsilon -1)=0$, that is $\epsilon - 1=0$.

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    $\begingroup$ Dear Suzet, I think you can just simplify the $i$ in your equation $i\circ\epsilon= i$ to obtain the result. Imho a simpler way to show the above is using the explicit description (4.3.5) which amounts to explaining the action of $\operatorname{Aut}(E/S)$ on the set of homs and then applying the description. (In essence the same thing). $\endgroup$ – user1728 Aug 13 '18 at 16:20

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