What spaces $X$ do have $\text{End}(X) \cong \text{End}(\mathbb{R})$? This is a follow-up on the following question. Let $\text{End}(X)$ denote the endomorphism monoid of a topological space $X$ (that is, the collection of all continuous maps $f:X\to X$ with composition).
What is an example of a topological space $X$ with $X\not\cong \mathbb{R}$ but the monoids $\text{End}(X)$ and $\text{End}(\mathbb{R})$ are isomorphic?
 A: No such space exists. We actually get the stronger statement that every isomorphism $\operatorname{End}(X) \stackrel\sim\to \operatorname{End}(\mathbf R)$ is induced by an isomorphism $X \stackrel\sim\to \mathbf R$ (unique by Observation 1 below). In contrast, in Emil Jeřábek's beautiful construction in this parallel post there is an 'outer automorphism' $\operatorname{End}(X) \stackrel\sim\to \operatorname{End}(X)$ that does not come from an automorphism $X \stackrel\sim\to X$ of topological spaces (it comes from an anti-automorphism of ordered sets).
I will use the substantial progress by YCor and Johannes Hahn, summarised as follows:
Observation 1 (YCor). For every topological space $X$, the map $X \to \operatorname{End}(X)$ taking $x$ to the constant function $f_x$ with value $x$ identifies $X$ with the set of left absorbing¹ elements of $\operatorname{End}(X)$.
In particular, an isomorphism of monoids $\operatorname{End}(X) \stackrel\sim\to \operatorname{End}(Y)$ induces a bijection $U(X) \stackrel\sim\to U(Y)$ on the underlying sets.
Observation 2 (Johannes Hahn). If $\operatorname{End}(X) \cong \operatorname{End}(\mathbf R)$, then $X$ is $T_1$. Since the closed subsets of $\mathbf R$ are exactly the sets of the form $f^{-1}(x)$ for $x \in \mathbf R$, we conclude that these are closed in $X$ as well, so the bijection $X \to \mathbf R$ of Observation 1 is continuous.
(The asymmetry is because we used specific knowledge about $\mathbf R$ that we do not have about $X$.)

To conclude, we prove the following lemma.

Lemma. Let $\mathcal T$ be the standard topology on $\mathbf R$, and let $\mathcal T' \supsetneq \mathcal T$ be a strictly finer topology. If all continuous maps $f \colon \mathbf R \to \mathbf R$ for $\mathcal T$ are continuous for $\mathcal T'$, then $\mathcal T'$ is the discrete topology.

Note that Observation 2 and the assumption $\operatorname{End}(X) \cong \operatorname{End}(\mathbf R)$ imply the hypotheses of the lemma, so we conclude that either $X = \mathbf R$ or $X = \mathbf R^{\operatorname{disc}}$. The latter is clearly impossible as it has many more continuous self-maps.
Proof of Lemma. Let $U \subseteq \mathbf R$ be an open set for $\mathcal T'$ which is not open for $\mathcal T$. Then there exists a point $x \in U$ such that for all $n \in \mathbf N$ there exists $x_n \in \mathbf R$ with $|x - x_n| \leq 2^{-n}$ and $x_n \not\in U$. Without loss of generality, infinitely many $x_n$ are greater than $x$, and we can throw out the ones that aren't (shifting all the labels, so that $x_0 > x_1 > \ldots > x$). Up to an automorphism of $\mathbf R$, we can assume $x = 0$ and $x_n = 2^{-n}$ for all $n \in \mathbf N$. Taking the union of $U$ with the usual opens $(-\infty,0)$, $(1,\infty)$, and $(2^{-n},2^{-n+1})$ for all $n \in \mathbf N$ shows that
$$Z = \big\{1,\tfrac{1}{2},\tfrac{1}{4},\ldots\big\}$$
is closed for $\mathcal T'$. Consider the continuous function
\begin{align*}
f \colon \mathbf R &\to \mathbf R\\
x &\mapsto \begin{cases}0, & x \leq 0,\\ x, & x \geq 1, \\ 2^nx, & x \in \big(2^{-2n},2^{-2n+1}\big], \\ 2^{-n}, & x \in \big(2^{-2n-1},2^{-2n}\big].\end{cases}
\end{align*}
Then $f^{-1}(Z)$ is the countable union of closed intervals
$$Z' = \bigcup_{n \in \mathbf N} \big[2^{-2n-1},2^{-2n}\big] = \big[\tfrac{1}{2},1\big] \cup \big[\tfrac{1}{8},\tfrac{1}{4}\big] \cup \ldots.$$
By the assumption of the lemma, both $Z'$ and $2Z'$ are closed in $\mathcal T'$, hence so is the union
$$Z'' = Z' \cup 2Z' \cup [2,\infty) = (0,\infty),$$
and finally so is $Z'' \cup (-Z'') = \mathbf R\setminus 0$. Thus $0$ is open in $\mathcal T'$, hence so is every point, so $\mathcal T'$ is the discrete topology. $\square$

¹Elements $f$ such that $fg = f$ for all $g$. (I would probably have called this right absorbing!)
A: By @YCor's comment, $X$ has the same number of elements as $\mathbb{R}$ and the actions of $End(X)$ on $X$ is the same as the action of $End(\mathbb{R})$ on $\mathbb{R}$. Now consider the automorphism group and its action on $X$. $Aut(\mathbb{R})$ consists of strictly increasing and strictly decreasing maps and it is easy to see that the point stabiliser subgroups $Aut(\mathbb{R})_x$ act transitively on $\mathbb{R}\setminus\{x\}$. But this action is imprimitive: $\mathbb{R}\setminus\{x\}$ has two blocks, namely $(x,+\infty)$ and $(-\infty,x)$ and the set stabilisers act transitively on those two sets. Phrased differently: There are exactly three equivalence relations on $\mathbb{R}\setminus\{x\}$ invariant under $Aut(\mathbb{R})_x$, the two trivial ones "everything is equivalent" and "nothing is equivalent" as well as a unique non-trivial one with the two equivalence classes $(-\infty,x)$ and $(x,+\infty)$.
Conclusion: We can recover a linear order on $X$ from $Aut(X)$ and thus from $End(X)$. And in particular we can recover the order topology of this ordering. And all elements of $End(X)$ must be continuous w.r.t. this order topology.
This does not necessarily mean that the order topology coincides with the original topology on $X$, but it is very close.

What can we say about the original topology? We know that there are lots of continuous map, but not too many (since $|End(\mathbb{R})|=|\mathbb{R}|$) so that $X$ is not indiscrete. We can say that the only self-maps with finite image are constant. In particular, there cannot be a Sierpinski space inside $X$, because every open set yields a continuous map to the Sierpinski space. Therefore $X$ is at least a $T_1$-space.
That is enough to conclude that the original topology on $X$ is at least as fine as the order topology: For every order-topology-closed subset $A\subseteq\mathbb{R}$ there is a continuous self-map $f:\mathbb{R}\to\mathbb{R}$ with $A=f^{-1}(0)$. Therefore the corresponding order-topology-closed subset of $X$ is also the preimage of a point under a continuous self-map and thus original-topology-closed. In other words: Our canonical bijection $X\to\mathbb{R}$ is continuous w.r.t. the original topology on $X$.
I have the feeling that the other direction is equally easy, but I just don't see it right now.
A: There is a straightforward approach to proving that isomorphism of monoids $\ \text{End}(X\ T)\ $ and
$\ \text{End}(\Bbb R\,\ T_E)\ $ implies homeomorphism of $\ (X\ T)\ $ and Euclidean space
$\ (\Bbb R\,\ T_E):\ $
let $\ (M\ \circ\ J)\ $ be an arbitrary abstract monoid. One may associate with this monoid a set of
"points" $\ C\ $ of "constant" elements, as @YCor did in his first comment, as the left-absorbing elements $\ c\in C,\ $ where
$$ \forall_{f\in M}\quad c\circ f=c $$
Then one selects purely algebraic (monoidal) notions which have the respective topological meaning in $\ (\Bbb R\,\ T_E).\ $ You need just a couple of such notions.
Then you force on the abstract monoid $\ (M\ \circ\ J)\ $ the purely algebraic (monoidal) axioms that induce the properties of reals. That's all.
For instance one may apply idempotents $\ i\in\mathcal I\subseteq M,\ $ where $\ i\circ i=i. \ $ Then the critical notion to me is what I've defined and call universally closed idempotents or uc-morphism $\ i\in\mathcal I,\ $ which satisfy:
$$  \forall_{f\in M}\,\exists_{p\in M}\quad
    i\circ f\circ i\circ p\,\ =\,\ i\circ p $$
Observe that we have a canonical map
$$ \pi: \mathcal I\to 2^C, $$
where
$$ \forall_{i\in\mathcal I}\quad \pi(i)\ :=
   \ \{c\in C:\ i\circ c=c\} $$
This has nice properties... etc.
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(After I saw the OP question, I hesitated... and I can remove my post, no sweat.)
