$\newcommand{\Fix}{\mathrm{Fix}}\newcommand{\N}{\mathbb{N}}\newcommand{\id}{\mathrm{id}}\newcommand{\End}{\mathrm{End}}\newcommand{\Tr}{\mathrm{Tr}}\newcommand{\tr}{\mathrm{tr}}$Here are some extra details and explicit computations for determining the trace $\Tr(\Delta)$ of $\Delta$ as well as the trace map
$$\tr\colon\End(\Delta)\to\Tr(\Delta),$$
following Maxime's answer.

**An Explicit Example**

Perhaps the general way to proceed is best exemplified by an example: consider the map $f\colon [4]\to[4]$ given by

and let's denote it (and maps of $\Delta$ in general) by the cleaner notation

To compute $\tr(f)$, we can write this map as the composition of a surjection $s\colon [4]\twoheadrightarrow[2]$ with an injection $i\colon[2]\hookrightarrow[4]$:

Now, since we have $\tr(f\circ g)=\tr(g\circ f)$ for any two endomorphisms $f$ and $g$ of $\Delta$ such that $f\circ g$ and $g\circ f$ are both defined, we see that $\tr(f)=\tr(i\circ s)=\tr(s\circ i)$, where $s\circ i$ is the map

which is the identity map $\id_{[2]}\colon[2]\to[2]$ of $[2]$ in $\Delta$.

**The General Case**

**Claim 1.** Let $f\colon[n]\to[n]$ be an arbitrary endomorphism of $\Delta$. We have $\tr(f)=\tr\big(\id_{[k]}\big)$ for some $k\leq n$.

*Proof.* If $f=\id_{[n]}$, then we are done. Otherwise, we can again decompose it into a surjection $s\colon[n]\twoheadrightarrow[k]$ followed by an injection $i\colon[k]\hookrightarrow[n]$, where $k=\#\mathrm{Im}(f)$ will be strictly less than $n$. By the property $\tr(f\circ g)=\tr(g\circ f)$ of the trace map, we have
\begin{align*}
\tr(f) &= \tr(i\circ s)\\
&= \tr(s\circ i),
\end{align*}
with $s\circ i\colon[k]\to[k]$ again an endomorphism of $\Delta$. Now, if $s\circ i=\id_{[k]}$, we are done. Otherwise, we repeat this process again.

Since $n$ is finite, it follows that repeating this process will give us a natural number $n_0\leq n$ such that $\tr(f)=\tr\big(\id_{[n_0]}\big)$ in finitely many steps.

Next, we aim to show the following:

**Claim 2.** If $n\neq m$, then $\tr\big(\id_{[n]}\big)\neq\tr\big(\id_{[m]}\big)$.

To this end, we want we want to construct a map
$$\phi\colon\mathrm{End}(\Delta)\to\N_{\geq1}$$
satisfying the following conditions:

- If $n\neq m$, then $\phi\big(\id_{[n]}\big)\neq\phi\big(\id_{[m]}\big)$.
- We have $\phi(f\circ g)=\phi(g\circ f)$, so that $\phi$ descends to a map $\widetilde{\phi}\colon\mathrm{Tr}(\Delta)\to\N_{\geq1}$.

From the example of $f\colon[4]\to[4]$ above and similar ones, we might guess that the map $\sigma\mapsto\#\mathrm{Im}(\sigma)$ sending an endomorphism of $\Delta$ to the cardinality of its image would be our desired map $\phi$.

However, as Peter pointed out, the map $\sigma\mapsto\#\mathrm{Im}(\sigma)$ fails to satisfy Item (2) above, as the following example shows:

Consider the maps $f\colon[3]\to[2]$ and $g\colon[2]\to[3]$ given by

Then the maps $g\circ f\colon[3]\to[3]$ and $f\circ g\colon[2]\to[2]$ are given by

and we have $\#\mathrm{Im}(g\circ f)=2$ while $\#\mathrm{Im}(f\circ g)=1$.

The map sending $\sigma\colon[n]\to[n]$ to the cardinality $\#\Fix(\sigma)$ of its set of fixed points does fill both criteria:

If $n\neq m$, then $\#\Fix(\id_{[n]})=n+1\neq m+1=\#\Fix(\id_{[m]})$.

We have $\#\Fix(g\circ f)=\#\Fix(f\circ g)$, as the maps
\begin{align*}
f|_{\Fix(g\circ f)} &\colon \Fix(g\circ f) \to \Fix(f\circ g),\\
g|_{\Fix(f\circ g)} &\colon \Fix(f\circ g) \to \Fix(g\circ f)
\end{align*}
set up a bijection between $\Fix(g\circ f)$ and $\Fix(f\circ g)$, as:

a) If $x$ is a fixed point of $g\circ f$, i.e. $g(f(x))=x$, then $f(x)$ is a fixed point of $f\circ g$, since $f(g(f(x))=f(x)$.

b) Similarly, if $y$ is a fixed point of $f\circ g$, then $g(y)$ is a fixed point of $g\circ f$.

c) For each $x\in\Fix(g\circ f)$, we have
\begin{align*}
[g|_{\Fix(f\circ g)}\circ f|_{\Fix(g\circ f)}](x) &= g(f(x))\\
&= x\\
&= [\id_{\Fix(g\circ f)}](x).
\end{align*}
so $g|_{\Fix(f\circ g)}\circ f|_{\Fix(g\circ f)}=\id_{\Fix(g\circ f)}$.

d) Similarly, $f|_{\Fix(g\circ f)}\circ g|_{\Fix(f\circ g)}=\id_{\Fix(f\circ g)}$.

It follows that $\sigma\mapsto\#\Fix(\sigma)$ defines a map $\widetilde{\phi}\colon\Tr(\Delta)\to\mathbb{N}_{\geq1}$.

**Finally, the result.**

Since:

- For all $\tr(f)\in\Tr(\Delta)$, there exists some $n\in\mathbb{N}_{\geq1}$ such that $\tr(f)=\tr\big(\id_{[n]}\big)$ (Claim 1).
- If $n\neq m$, then $\tr\big(\id_{[n]}\big)\neq\tr\big(\id_{[m]}\big)$ (Claim 2).

it follows that $\widetilde{\phi}$ is a bijection:

*$\widetilde{\phi}$ Is Injective:* Given $\tr(f)$ and $\tr(g)$ with $\widetilde{\phi}(\tr(f))=\widetilde{\phi}(\tr(g))$, there exist $n,m\in\mathbb{N}_{\geq1}$ such that $\tr(f)=\tr\big(\id_{[n]}\big)$ and $\tr(g)=\tr\big(\id_{[m]}\big)$ by Claim 1, but since $\widetilde{\phi}(\tr(f))=\widetilde{\phi}(\tr(g))$, we must have $n=m$ by Claim 2, so $\tr(f)=\tr(g)$.
*$\widetilde{\phi}$ Is Surjective:* Given $n\in\N_{\geq1}$, we have $\widetilde{\phi}\big(\tr\big(\id_{[n-1]}\big)\big)=n$.

So $\Tr(\Delta)\cong\mathbb{N}_{\geq1}$. Lastly, to determine the trace map

$$\tr\colon\End(\Delta)\to\Tr(\Delta),$$

note that we have $\phi=\tr\circ\widetilde{\phi}$ by the universal property of the quotient, and since $\widetilde{\phi}$ is a bijection, we have $\tr=\widetilde{\phi}{}^{-1}\circ\phi$, which is given by $\sigma\mapsto\tr\big(\id_{[\#\Fix(\sigma)-1]}\big)$.