Here is a sort of partial solution. I doubt it will be very helpful, but anyone who wants to read it is free to do so.
Let $I$ be any space with the fixed point property. We will construct a space $X$ and a map $X \to I^X$ such that many interesting maps lie in its image, including all maps that can be defined in a language consisting of $e$, continuous functions from finite powers of $I$ to itself, and constant symbols in $X$ (which are all that is needed to prove the Lawvere fixed point theorem).
To accomplish this we will inductively construct for each natural number $n$ a space $X_n$, a map $e_n: X_n \times X_n \to I$, and a map $i_n: X_n \to X_{n+1}$, such that $e_{n+1}(i_n(x_1),i_n(x_2))=e_n(x_1,x_2)$. We will also use auxiliary spaces $Y_n,Z_n$ along the way.
After this we set $X$ to be the forward limit of $X_n$ along $i_n$ and let $e: X\times X \to I$ be the limit of the $e_n$.
The significance of the spaces $Y_n$ and $Z_n$ is that $Z_n$ is the set of continuous $I$-valued functions of $x \in X$ that depend only on $e(t,x), e(x,x), e(x,t)$ for $t \in X_n$. $Y_n$ is the space of possible values of the triple $e(t,x), e(x,x), e(x,t)$, so that $Z_n$ is the space of functions on $Y_n$. Then $X_{n+1}$ is constructed so that it maps surjectively to $Z_n$, so all functions of $Z_n$ come from elements of $X_n$, and maps to $Y_n$, so that all functions of $Z_n$ can be extended to functions on $X_n$. For simplicity and canonicality, we define $X_{n+1}$ to be a subset of $Y_n \times Z_n$, defined by coherence condition to ensure the desired relationship between $Y_n$ and $e$.
To begin, let $X_0$ be the empty set.
Inductively, assume we have defined $X_{n-1}, Y_{n-1},Z_{n-1}, X_n$.
Let $Y_n$ be the subset of $I^{X_n} \times I \times I^{X_n}$ consisting of tuples (a,b,c) satisfying the coherence conditions:
$(a \circ i_{n-1} , b, c \circ i_{n-1}) \in Y_{n-1}$
For all $x$ in $X_{n}$ which corespond to a tuple $(a',b',c'),f'$ in $Y_{n-1} \times Z_{n-1}$, $a(x)= f'( a \circ i_{n-1} ,b, c \circ i_{n-1})$.
In the case $n=0$, so $X_n = \emptyset$, this is vacuous, and so we need not introduce $Y_{-1}$ and $Z_{-1}$.
Let $Z_n= I^{Y_n}$.
Let $X_{n+1} $ be the subset of $Y_n \times Z_n$ consisting of tuples $(a,b,c)\in Y_n, f\in Z_n$ satisfying the following two coherence conditions:
$b= f(a,b,c)$
For all $x$ in $X_n$, $c(x) = f( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t))$
Let $i_n: X_n \to X_{n+1}$ send $x$ to $( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t) ) , f(\alpha,\beta,\gamma)=\alpha(x)$.
Let $e_{n+1}: X_{n+1} \times X_{n+1} \to I$ send $((a_1,b_1,c_1),f_1),((a_2,b_2,c_2),f_2)$ to $f_1(a_2,b_2,c_2)$.
Let us first check that for $x$ in $X_n$ corresponding to a tuple $(a,b,c),f$ in $Y_{n-1} \times Z_{n-1}$, $i_n(x)$ satisfies the coherence conditions.
For $x'$ in $X_n$ which corresponds to a tuple $(a',b',c'),f'$, we need $e_n(x',x) = f'(t \mapsto e_n( i_{n-1}(t),x), e_n(x,x), t \mapsto e_n(x,i_{n-1}(t))$. So it is sufficient to check that $a(t) = e_n( i_{n-1}(t),x)$, $b=e_n(x,x)$, and $c=e_n(x,i_{n-1}(t))$, which follow from the definition of $e_n$ and respectively the definition of $i_{n-1}$, the coherence condition for $b$, and the definition of $i_{n-1}$ as well as the coherence condition for $c$.
Hence $i_n$ is actually a well-defined map from $X_n$ to $X_{n+1}$.
The fact that $e_{n+1}(i_n(x_1),i_n(x_2))=e_n(x_1,x_2)$ follows from unwinding the definitions.
Now we can define $X$ to be the forward limit of $X_n$ along $I_n$ and $e$ to be the forward limit of the maps $e_n$, which we have seen are compatible. Next we will characterize the image of the map $X \to I^X$ given by $x \mapsto (t \mapsto e(x,t))$. We will see that it this image can be viewed as the forward limit of $Z_n$ along the system of maps that we now define.
Consider the map $j_n: Y_{n+1} \to Y_{n}$ that sends $(a,b,c)$ to $(a \circ i_n, b, c \circ i_n)$. Consider also the map $k_n: Z_{n} \to Z_{n+1}$ by exponentiating $j_n$.
I claim that if $i_{n+1} ((a,b,c),f)= ((a',b',c'),f')$ then $j_n(a',b',c') = (a,b,c)$ and $k_n(f)=f'$.
The first statement is simply the fact that $ a(t)= t \mapsto e_n(i_{n-1}(t),x), b=e_n(x,x), c(t)= t \mapsto e_n(x,i_{n-1}(t))$, the first of which follows from the definition of $e_n$ and the second and third from the definition of $e_n$ and the coherence conditions for $b$ and $c$ respectively.
The second statement is simply the statement that for $ (\alpha,\beta,\gamma)$ in $Y_{n+1}$, $f( \alpha \circ i_n, \beta , \gamma \circ i_n) = \alpha(x)$ which is the coherence condition for $a$ applied to $\alpha$.
So we have verified the claim.
Let $Y$ be the inverse limit of $Y_n$ along $j_n$ and let $Z$ be the inverse limit of $Z_n$ along $k_n$, so that there is a natural map $Y \times Z \to I$. The compatibilities of $i_n$ with $j_n$ and $k_n$ respectively imply that there are map $X \to Y$ and $X \to Z$, and $e: X \times X \to I$ is simply the composition of these two with the map $Y \times Z \to I$.
Hence the map $X \to I^X$ induced by $e$ factors as $X \to Z \to I^Y \to I^X$. I can't prove that this composition is surjective but I can prove that the first map, $X \to Z$ is surjective.
To do this, it is sufficient to prove that the projection map $X_n \to Z_n$ is surjective. In other words, given $f: Y_n \to I$, construct $(a,b,c) \in Y_n$ satisfying the coherence conditions. Clearly we must have $c(x) = f( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t))$. For now, let $b$ be a parameter. Define $c_m: X_m \to I$ for $m$ from $0$ to $n$ such that $c_n= c$, $c_m = c_{m+1} \circ i_m$, then define $a_m: X_m \to I$ so that $a_0$ is the empty function and $a_m(x)$ for $x \in X_m$ corresponding to $(a',b',c',f')$ is $f'(a_{m-1}, b, c_{m-1})$. Then by the definition of $i_{m-1}$, $a_{m-1} =a_m \circ i_{m-1}$. Now each $a_m$ is a continuous function of the previous, so $a_n$ is a continuous function of $b$, hence $b\mapsto f(a_n,b,c)$ has a fixed point. So we may take $a=a_n$, $b$ to be the fixed point and $c=c$.
What maps can we construct this way? We get all the maps defined by elements of $Z_n$ for all $n$. These, in turn, are all continuous maps that depend only on the data in $Y_n$. By the coherence conditions and the definition of $e$, the projection of $x \in X_n$ to $Y_n$ encode $t\mapsto e(t,x),e(x,x),t\mapsto e(x,t)$ for $t$ in $X_{n-1}$. The construction of the projection $j_n: Y_{n+1} to Y_n$ ensure that it continues to encode this data for all $x$. Hence all continuous functions that depend only on $t\mapsto e(t,x),e(x,x),t\mapsto e(x,t)$, which seem to me to be all the "obvious" ones that we construct, are in the image of the constructed map $X \to I^X$.
The definability statement from earlier follows from the fact that any finite set of elements of $X$, such as the constant symbols appearing in the formula, must lie in $X_n$ for some $n$. Any function that depends only on $e$ evaluated with these constant symbols will then lie in $Z_n$.