Let $C$ be a category composed by three pieces: $C_{-2}, C_0$ and $C_2$, where $C_{-2}=C_2=\mathbb{K}-$modules and $C_0 = \frac{\mathbb{K}[x]}{(x^2)}$-modules.
Let $E$ be the functor that acts this way: $E: C_{-2} \xrightarrow{Ind} C_0, C_0 \xrightarrow{Res} C_2$ , $0$ otherwise.
My book then says that, named $T \in \operatorname{End}_\mathbb{K}\left(\frac{\mathbb{K}[x]}{(x^2)}\right)$ that swaps $1$ and $x$, $T$ gives us an endomorphism $\tilde{T} \in \operatorname{End}(E^2)$.
1) How is $T$ an endomorphism in the first place? I think it's not even well-defined, I mean, I get $$1=T(x) = T(x\cdot 1 \cdot 1) = 1 \cdot x \cdot x = x^2 = 0$$
2) I believe I really don't get how "endomorphism of functors" work, because in my computations $E^2$ should basically be a functor from $C_{-2}$ to $C_2$ (it's supposed to be $0$ anywhere else). So, even if we had an actual $T$, how would that give us an element of $\operatorname{End}(E^2)$ ?
I figured it out on my own, so since one of you $\star$ed the question I thought it would be nice to share (and, well, it may still be wrong so another pair of eyes would be nice too)
1) We are viewing it as an endomorphism of vector space, so it just has to be $\mathbb{K}$ linear. My example then is just wrong, since as a vector space $\frac{\mathbb{K}[x]}{(x^2)} \simeq \mathbb{K}^2$, generated by $v=1$ and $w=x$. Swapping the two generators obviously gives us an endomorphism
2) We just need to write down what $E^2 V$ is, for some $V\in C_{-2}$... $E^2$ is indeed $0$ on the other two pieces.
So, let $V$ be a $\mathbb{K}$-module (a vector space). By definition of the induction functor, $$EV = V \otimes_{\mathbb{K}} \frac{\mathbb{K}[x]}{(x^2)}$$ where the action of $\frac{\mathbb{K}[x]}{(x^2)}$, because of $\mathbb{K}$ linearity, needs only to be given on $1 \otimes 1$ and $1 \otimes x$, and we define that as $$(\beta x).(1 \otimes 1) = \beta (1 \otimes x)$$ $$(\beta x).(1 \otimes x) = 0$$
Now, by definition of the restriction functor, given any $M \in C_0$ (i.e. a $\frac{\mathbb{K}[x]}{(x^2)}$-module ), $EM$ is just the same $M$ seen as a $\mathbb{K}$-module, so we are just forgetting that $x$ can act on it, and look at it as a vector space.
So $E^2V$ is just the same $V \otimes_{\mathbb{K}} \frac{\mathbb{K}[x]}{(x^2)}$, seen as a vector space.
Now it makes sense to define an endomorphism $T_M$ of such an object, just putting $$ T_M(1 \otimes 1) = 1 \otimes x \quad , \quad T_M(1 \otimes x) = 1 \otimes 1$$
Now, is this an endomorphism of functors? We need to check that for any $M, M' \in C_{-2}$, and for any $f: M \to M'$ $$E^2 f \circ T_M = T_M \circ E^2 f$$ Defining $E^2 f$ in the obvious way
$\big($ $Ef$ is just the natural $Ef: M \otimes \frac{\mathbb{K}[x]}{(x^2)} \to M' \otimes \frac{\mathbb{K}[x]}{(x^2)}$ that sends $m \otimes n$ to $f(m) \otimes n$$\big)$
it is clear that swapping the "right parts" of the tensor doesn't involve $f$, viceversa. In other words $$(E^2 f \circ T_M )(m \otimes (ax+b))= E^2 f (m \otimes (bx+a)) = \\ f(m) \otimes (bx+a) \\ = T_M (f(m) \otimes ax+b) = (T_M \circ E^2f)(m \otimes (ax+b))$$ and we have the desired element $T \in \operatorname{End}(E^2)$.