$\DeclareMathOperator\End{End}$As shown in Todd Trimble’s comment, the set of constant maps $X\to X$ is definable in $\End(X,\tau)$, as it consists of exactly the left-absorbing endomorphisms (i.e., $\phi\in\End(X,\tau)$ such that $\phi\circ\psi=\phi$ for all $\psi\in\End(X,\tau)$). Thus, an isomorphism $F\colon\End(X,\tau)\to\End(Y,\sigma)$ induces a bijection $f\colon X\to Y$ such that $F(c_x)=c_{f(x)}$ for all $x\in X$, where $c_x\colon X\to X$ is the constant-$x$ map. But then $F$ is completely determined by $f$ by
$$F(\phi)(f(x))=f(\phi(x))$$
for all $\phi\in\End(X,\tau)$: indeed, we have
$$c_{f(\phi(x))}=F(c_{\phi(x)})=F(\phi\circ c_x)=F(\phi)\circ F(c_x)=F(\phi)\circ c_{f(x)}=c_{F(\phi)(f(x))}.$$
Since we may as well assume that $X=Y$ and $f$ is the identity, it follows that:
Lemma 1. $(X,\tau)$ has a unique endomorphism monoid iff it is homeomorphic to all spaces of the form $(X,\sigma)$ such that $\End(X,\tau)$ and $\End(X,\sigma)$ are literally equal (i.e., a map $X\to X$ is an endomorphism of $(X,\tau)$ iff it is an endomorphism of $(X,\sigma)$.)
This implies
Proposition 2. If $(X,\le)$ is a total order which is isomorphic to its opposite order, then the Alexandrov space $(X,\tau)$ of upper sets of $(X,\le)$ has a unique endomorphism monoid. In particular, there exist such spaces of arbitrary cardinality.
Indeed, let $\sigma$ be a topology on $X$ such that $\End(X,\sigma)$ consists of the order-preserving maps. We may assume $|X|\ge2$. Then $\sigma$ cannot be indiscrete, hence we can fix $V\in\sigma$ and $a$ and $b$ such that $a\notin V$, $b\in V$. Assume first $a<b$. Then $\tau\subseteq\sigma$: consider an upper set $U\in\tau$. The map
$$\phi_{a,b,U}(x)=\begin{cases}b&x\in U,\\a&x\notin U\end{cases}$$
is order-preserving, hence $\phi_{a,b,U}\in\End(X,\sigma)$, and $\phi_{a,b,U}^{-1}[V]=U$, thus $U\in\sigma$. In fact, $\sigma=\tau$: if we assume for contradiction that $W\in\sigma$ is not an upper set, then the argument above shows that $\sigma$ also includes all lower sets, thus for any upper set $U$, $\phi_{b,a,U}\in\End(X,\sigma)$, but $\phi_{b,a,U}$ is not order-preserving if $U\notin\{\varnothing,X\}$.
Dually, if $a>b$, we obtain that $\sigma$ consists of all lower subsets of $X$, hence it is the Alexandrov topology corresponding to the opposite of $\le$, but this is homeomorphic to $(X,\tau)$ by our assumption on $\le$.
One can generalize the argument to a complete characterization for Alexandrov spaces. (Note that in particular, all finite spaces are Alexandrov.) First, a lemma. If $(X,\tau)$ is a topological space, let $x\le_\tau y$ denote the specialization preorder $x\in\overline{\{y\}}$, and $x\sim_\tau y$ the indistinguishability equivalence $x\le_\tau y\land y\le_\tau x$.
Lemma 3. If $\End(X,\tau)=\End(X,\sigma)$, then ${\sim_\tau}={\sim_\sigma}$ unless one space is discrete and the other indiscrete.
Proof: If, say, $a\sim_\tau b$ but $a\nsim_\sigma b$, then all mappings $X\to\{a,b\}$ are in $\End(X,\tau)$, hence in $\End(X,\sigma)$, hence $(X,\sigma)$ is discrete, hence all mappings $X\to X$ are in $\End(X,\tau)$, hence $(X,\tau)$ is indiscrete lest $a\nsim_\tau b$. QED
Proposition 4. If $(X,\tau)$ is an Alexandrov space, then $\End(X,\tau)\simeq\End(Y,\sigma)$ if and only if
$(Y,\sigma)$ is homeomorphic to $(X,\tau)$, or
$(Y,\sigma)$ is homeomorphic to the opposite of $(X,\tau)$ (i.e., the Alexandrov space corresponding to $\ge_\tau$), or
the two spaces are the discrete and indiscrete topologies on sets of the same cardinality.
Consequently, $(X,\tau)$ has a unique endomorphism monoid iff $(X,\le_\tau)\simeq(X,\ge_\tau)$, and $\tau$ is neither discrete nor indiscrete unless $|X|\le1$.
Proof: The right-to-left implication is clear. For the left-to-right implication, we may assume $X=Y$ and $\End(X,\tau)=\End(X,\sigma)$ as above.
Assume first that $\le_\tau$ is an equivalence (i.e., ${\le_\tau}={\sim_\tau}$). By Lemma 3, we may assume that ${\sim_\tau}={\sim_\sigma}$ and $\tau$ is not indiscrete. (If $\tau$ is indiscrete, then either $\sigma$ is discrete and we are done, or ${\sim}_\sigma={\sim}_\tau$, hence $\sigma$ is indiscrete, i.e., $\sigma=\tau$, and we are also done.) Since $\tau$ is the finest topology with indistinguishability relation $\sim_\tau$, this implies $\sigma\subseteq\tau$; on the other hand, if we fix $a\nsim_\tau b$, and w.l.o.g. $a\lnsim_\sigma b$, then $\phi_{a,b,U}$ is in $\End(X,\sigma)$ for all $U\in\tau$, hence $U\in\sigma$, i.e., $\sigma=\tau$.
If $\le_\tau$ is not an equivalence, let us fix $a\lnsim_\tau b$. This also implies we can fix $V\in\tau$ whose complement is not in $\tau$. Then $\phi_{a,b,V}\in\End(X,\tau)$ and $\phi_{b,a,V}\notin\End(X,\tau)$, hence $a\lnsim_\sigma b$ or $b\lnsim_\sigma a$. W.lo.g., we assume the former (the other choice leads to the opposite order). Then for each $U\in\tau$, $\phi_{a,b,U}\in\End(X,\tau)$ implies $U\in\sigma$, i.e., $\tau\subseteq\sigma$. Since $\tau$ is the finest topology with specialization preorder $\le_\tau$, if $\tau\subsetneq\sigma$, then (in view of ${\sim_\tau}={\sim_\sigma}$) there are $x,y$ such that $x\lnsim_\tau y$ and $x\nleq_\sigma y\nleq_\sigma x$. But as above, this contradicts $\phi_{x,y,V}\notin\End(X,\sigma)$ for suitable $V\in\tau$. Thus, $\tau=\sigma$. QED
The characterization can be easily extended to all non-$R_0$ spaces. Recall that $(X,\tau)$ is $R_0$ if $\le_\tau$ is symmetric (i.e., ${\le_\tau}={\sim_\tau}$).
Proposition 5. If $(X,\tau)$ is a non-Alexandrov non-$R_0$ space, then $(X,\tau)$ has a unique endomorphism monoid.
Proof: Assume that $\End(X,\tau)=\End(X,\sigma)$. Let us fix $a\lnsim_\tau b$. There exists $V\in\tau$ whose complement is not in $\tau$ (e.g., any open set separating $b$ from $a$); then $\phi_{a,b,V}\in\End(X,\tau)$ and $\phi_{b,a,V}\notin\End(X,\tau)$, hence (1) $a\lnsim_\sigma b$ or (2) $b\lnsim_\sigma a$. (In particular, $(X,\sigma)$ is not $R_0$.) If (1) holds, then for every $U\in\tau$, $\phi_{a,b,U}\in\End(X,\tau)$ implies $U=\phi_{a,b,U}^{-1}[b]\in\sigma$, i.e., $\tau\subseteq\sigma$. If (2) holds, then the same argument gives $\{X\smallsetminus U:U\in\tau\}\subseteq\sigma$.
Since $(X,\sigma)$ is not $R_0$ either, a symmetric argument implies that (1') $\sigma\subseteq\tau$, or (2') $\{X\smallsetminus U:U\in\sigma\}\subseteq\tau$. It is impossible that (1) and (2') hold together: this would imply that $\tau$ is closed under complement, whence it is $R_0$. Likewise, (2) and (1') are incompatible. Thus, the only two possibilities are that either (1) and (2) hold, in which case $\tau=\sigma$, or (1') and (2') hold, in which case $\tau$ and $\sigma$ are mutually opposite Alexandrov spaces. QED
Notice that $(X,\tau)$ is $R_0$ iff the Kolmogorov quotient $X/{\sim_\tau}$ is $T_1$. It is easy to see that:
Lemma 6. If $(X,\tau)$ and $(X,\sigma)$ are spaces such that ${\sim_\tau}={\sim_\sigma}$, then $\End(X,\tau)=\End(X,\sigma)$ iff $\End((X,\tau)/{\sim_\tau})=\End((X,\sigma)/{\sim_\sigma})$.
In view of Lemma 3, this gives a reduction of the remaining classification to $T_1$ spaces. Observe that an $R_0$ space $(X,\tau)$ is Alexandrov iff $(X,\tau)/{\sim_\tau}$ is discrete.
Corollary 7. If $(X,\tau)$ is an $R_0$ non-Alexandrov space, then $(X,\tau)$ has a unique mononorphism monoid iff the $T_1$ space $(X,\tau)/{\sim_\tau}$ has a unique monomorphism monoid.
