I'd say that there is nothing algebraically that forces \(f^t(p)=p\) for every \(t\in\mathbb R\) when \(p\) is a fixed point of \(f^1\).
If I understand what you are asking, you are asking if the following conjecture is a theorem
Conjecture: let \(f:X\to X\) be a function and \(f(p)=p\) a fixed point. Then, given a way to non-integer iterating \(f\) using a superfucntion, i.e. a function \(\varphi(t,x)\) s.t. \(\varphi(0,x)=x\) and \(\varphi(t+1,x)=f(\varphi(t,x))\), we have that the orbit of the fixedpoint \(p\) is constant
\[\varphi(t,p)=p\]
I'd say there is nothing forcing this, at least algebraically.
First of all remember that we can build the superfunction piecewise, a la Andrew Robbins, obtaining k-differentiability on the endpoints.
We can define \(\Phi(n):=p\), the fixed point for every integer \(n\in\mathbb N\), then define \(\Phi(t)\) for \(t\in [0,1]\) to be an arbitrary loop in \(\gamma:[0,1]\to X\) that connects \(p\) to itself. You can then evaluate \[\Phi(x):=f^{{\rm floor}(x)}(\gamma (\{x\})) \].
Note that the constant solution \(\forall x.\,K(x)=p\) is a solution of the superfunction equation.
Maybe only adding constraints can force the orbit, the noodle, that connects the fixed point to itself at different iteration height, to be constant/straight line.
If I understand what you are asking, you are asking if the following conjecture is a theorem
Conjecture: let \(f:X\to X\) be a function and \(f(p)=p\) a fixed point. Then, given a way to non-integer iterating \(f\) using a superfucntion, i.e. a function \(\varphi(t,x)\) s.t. \(\varphi(0,x)=x\) and \(\varphi(t+1,x)=f(\varphi(t,x))\), we have that the orbit of the fixedpoint \(p\) is constant
\[\varphi(t,p)=p\]
I'd say there is nothing forcing this, at least algebraically.
First of all remember that we can build the superfunction piecewise, a la Andrew Robbins, obtaining k-differentiability on the endpoints.
We can define \(\Phi(n):=p\), the fixed point for every integer \(n\in\mathbb N\), then define \(\Phi(t)\) for \(t\in [0,1]\) to be an arbitrary loop in \(\gamma:[0,1]\to X\) that connects \(p\) to itself. You can then evaluate \[\Phi(x):=f^{{\rm floor}(x)}(\gamma (\{x\})) \].
Note that the constant solution \(\forall x.\,K(x)=p\) is a solution of the superfunction equation.
Maybe only adding constraints can force the orbit, the noodle, that connects the fixed point to itself at different iteration height, to be constant/straight line.
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)