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(x,x), e(x,t)$ for $t \in X_n$. $Y_n$ is the space of possible values of the pair $e(x,x), e(x,t)$, 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 $e$.  Then we can freely add also the functions $e(t,x)$ for $t\in X_n$ to $Z_n$, as they are already continuous functions of $e(x,x)$, $e(x,t)$ by construction of $X_n$.



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 $ I \times I^{X_n}$. Let $Z_n= I^{Y_n}$.

Let $X_{n+1} $ be the subset of $Y_n \times Z_n$ consisting of tuples $(b,c)\in Y_n, f\in Z_n$ satisfying the following two coherence conditions:

* $b= f(b,c)$

* For all $x$ in $X_n$, $c(x) = f(  e_n(x,x), t \mapsto e_n(x,t))$

Let $i_n: X_n \to X_{n+1}$ send $x= ((b',c'),f)$ to $(e_n(x,x), t \mapsto e_n(x,t) ) , (\beta,\gamma)\mapsto f'(\beta, \gamma \circ i_{n-1} )$.

Let $e_{n+1}: X_{n+1} \times X_{n+1} \to I$ send $((b_1,c_1),f_1),((b_2,c_2),f_2)$ to $f_1(b_2,c_2)$.

--------

Let us first check that for $x$ in $X_n$ corresponding to a tuple $(b,c),f$ 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 $(b,c),f$, we need $e_n(x,x) = f( e_n(x,x), t \mapsto e_n(x, i_{n-1}(t))$, but $e_n(x)=f(b,c)$ so it is sufficient to check that $b=e_n(x,x)$ and $c= t \mapsto e_n(x, i_{n-1}(t))$, which are the coherence conditions of $X_n$.

Furthermore we need for $x'$ in $X_n$ corresponding to a tuple $(b',c'),f)$, $e_n(x,x') = f( e_n(x',x'),t \mapsto e_n(x',i_{n-1}(t)))$, which is true because by definition of $e_n$, $e_n(x,x') = f(b',c')$ and $b' = f'(b',c') =e_n(x',x')$ while $c'(t)  =  f'( e_{n-1}(t,t), s \mapsto e_{n-1}(s,t)) = e_n(x', i_{n-1}(t))$ by definition of $e_n$ and $i_{n-1}$. 


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.

Indeed, let $x_1 = ((b_1,c_1),f_1)$ and let $f_2=((b_2,c_2),f_2))$ and then  $e_{n+1} (i_n ( x_1),i_n(x_2)) = f_1 ( e_n (x_2,x_2),  t \mapsto e_n(x_2, i_{n-1}(t)) =f_1(b_2,c_2)= e_n(x_1,x_2)$ by the coherence conditions for $b_2,c_2$ and the definition of $e_n$, and then the definition of $e_n$.

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 $(b,c)$ to $( b, c \circ i_n)$. Consider also the map $k_n: Z_{n} \to Z_{n+1}$ by exponentiating $j_n$. 

I claim that if $i_{n} ((b,c),f)= ((b',c'),f')$ then $j_n(b',c') = (b,c)$ and $k_n(f)=f'$.

The first statement is simply the fact that $  b=e_n(x,x), c(t)= t \mapsto e_n(x,i_{n-1}(t))$, which follow from the definition of $e_n$ and the coherence conditions for $b$ and $c$ respectively.

The second statement follows immediately from the construction of $i_n$, which forces $f' ( \beta,\gamma) = f( \beta , \gamma \circ i_n) = \alpha(x)$ which is exactly $f \circ k_n$.

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 $(b,c) \in Y_n$ satisfying the coherence conditions. Clearly we must have $c(x) = f(  e_n(x,x), t \mapsto e_n(x,t))$. Then we can take $b$ to be any fixed point of $f(b,c)$.

------------------

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 $e(x,x),t\mapsto e(x,t)$ for $t$ in $X_{n-1}$. Using the definition of $e$ and the compatibility of $e$ with $i_{n-1}$, one can see that it also encodes $t \mapsto e(t,x)$ for $t \in Y_{n-1}$. The compatibility of $i_n$ with $e$ and with $j_n$ ensure that $Y_n$ 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 $Z_n$.