Is it possible to set up multiple automorphisms over a structureless object inside single-sort defined category? I was trying to understand the behaviour of the primitive equality (=) in the axiomatization of category, which takes morphisms as primitives and objects as derivatives in bijection to identity morphisms, and based on the definitions I have found (which are all in the same vein), it seems that trying to set up two different automorphisms over one derived object A is impossible because they necessarily collapse to the identity morphism determining A via the idempotence axioms 1) and 2) below. This would mean that to introduce multiple automorphisms, which we need for example for groups characterized as one object categories, structure is essential. And if we want structure in purely categorical terms, we must move into a higher-order category, in order to enable this construction. So we might take a category of categories, where functors encode the structure of source into target in the obvious way.
Is there any way to set up the definition of category with derived objects so that there can exist multiple different automorphisms, and how would this be achieved? It might not be necessary for practical purposes, but as a consequence of this setup, the elementary equality check x ?= y seems to behave oddly due to axiom 5) - interacts with different levels of morphisms as though they were the same.
According to nLab definition https://ncatlab.org/nlab/show/single-sorted+definition+of+a+category#references
A category (single-sorted version) is a collection C, whose elements are called morphisms, together with two functions s,t:C→C and a partial function ∘:C×C→C, such that:

*

*s(s(x))=s(x)=t(s(x))

*t(t(x))=t(x)=s(t(x))

*x∘y is defined if and only if s(x)=t(y).

*If x∘y is defined, then s(x∘y)=s(y) and t(x∘y)=t(x).

*x∘s(x)=x and t(x)∘x=x (both composites are always defined, because of the first two axioms)

*(x∘y)∘z=x∘(y∘z), if either is defined (in which case the other is defined by the axiom 3).

 A: You might want to think about a concrete example: let's for a moment take $C$ to be a monoid. How would the 1-sorted definition work?
Let's be more precise: let $(M, e, \cdot)$ be a monoid (with carrier set $M$, identity element $e$ and binary operation $\cdot$). We want to create a category (and present it as a 1-sorted theory, as in the nlab page you linked) such that composition $\circ$ corresponds to the monoid operation $\cdot$.
First of all, notice how $x \cdot x$ is defined for all $x \in M$, so $s(x) = t(x)$ and hence (since our choice of $x$ was arbitrary) $s = t$: this is because of axiom $3$. But we also know that $x \cdot y$ is defined for all $x, y \in M$: this, together with the previous fact, forces $s(x) = s(y)$ (and likewise for $t$): the two function are also constant. To reiterate, all of this follows from axiom $3$ alone.
But at this point, axioms $1$ and $2$ are trivially satisfied, as well as axiom $4$. Axiom $6$ also holds, but this time because monoids are associative by definition.
The only thing left to figure out is what the (only) element $a$ in the image of $s$ (and hence, of $t$) is: to figure this out, we need to look at axiom $5$. After applying everything we already know about $s$ and $t$, axiom $5$ becomes
$$\forall x (x \cdot a = x = a \cdot x)$$
As you probably know, this is the defining property of the identity element $e$: we have now proven that $s(x) = t(x) = e$ for every $x \in M$ (up to proving that every monoid has a unique identity element, a well-known and easy-to-prove fact).
You might notice that never, in the discussion above, have we ever implied that all monoids are the trivial monoid $\{*\}$: in fact, the example we've been looking at is the prototypical 1-object category with (possibly) multiple automorphisms.
I can't tell what got you to think that if $s(f) = t(f)$ then $f$ is an identity; this is of course not the case. I can only guess: maybe you thought that if $s(f) = t(f)$, the fact that axiom $1$ and $2$ become the same statement somehow implies that $s(f) = t(f) = f$; if that's the case, it clearly isn't true.
Hopefully this clears things up a bit.
