Continuous-time extension of a discrete dynamical system It is clear that one can obtain a discrete dynamical system from a continuous one, but is the converse possible if the system is "nice"?
Define the discrete-time dynamical system on $\mathbb{R}^d$ by
$$
x_{n+1} = f(x_n);\, x_0\triangleq x
$$ 
where
$f \in C^2(\mathbb{R}^d;\mathbb{R}^d)$ and $x \in \mathbb{R}^d$.  
Fix a (large) positive integer $N$, is there a function $F:\mathbb{R}^d\rightarrow \mathbb{R}^d$ such that the solution to the continuous-time dynamical system
$$
\partial_t X_t = F(X_t) ; \qquad X_0=x,
$$
and $X_n = x_n$ for every $n \in \left\{1,\dots,N\right\}$?
 A: I provide a discrete expansion of an iterated function, but a symmetry constraint is also needed to simplify it to a continuous solution. Works for Schroeder's and Abel's Functional Equations. Note that convergence issues make if difficult to provide a complete answer.
Let $f(x)$ and $g(x)$ be functions in Banach space, then the composite $f(g(x))$ can be constructed using Faa Di Bruno's formula. 
\begin{eqnarray}
D^nf(g(x)) = \sum_{\pi(n)} \frac{n!}{k_1! \cdots k_n!} 
(D^kf)(g(x))
\left(\frac{Dg(x)}{1!}\right)^{k_1} \cdots
\left(\frac{D^ng(x)}{n!}\right)^{k_n}
\label{eq:FaaDiBruno}
\end{eqnarray}
where $\pi(n)$ denotes a partition of $n$, usually denoted by $1^{k_1}2^{k_2}\cdots n^{k_n}$ with $k_1+2k_2+ \cdots nk_n=k$; where $k_i$ is the number of parts of size $i$. The partition function $p(n)$ is a decategorized version of $\pi(n)$, the function $\pi(n)$ enumerates the integer partitions of $n$, while $p(n)$ is the cardinality of the enumeration of $\pi(n)$. [Comtet], [Riordan]
In turn $f(g(x))$ can construct any iterated function at a fixed point by setting $f(0)=0$ and $g(x)=f^{m-1}(x)$. 
\begin{equation}
    D^n f^t(x)= \sum_{\pi(n)} \frac{n!}{k_1! \cdots k_n!} 
(D^k f)(f^{t-1}(x))
\left(\frac{Df^{t-1}(x)}{1!}\right)^{k_1} \cdots
\left(\frac{D^n f^{t-1}(x)}{n!}\right)^{k_n}
\end{equation}
\begin{equation}
D^n f^t(0)=\sigma(n) + D'f(0) D^n f^{t-1}(0) 
\end{equation}
DynamicalRecurranceEquation
The Taylors series of $f^t(x)$ is derived by evaluating
the derivatives of the iterated function at a fixed point 
$f^t(0)$ by setting $x=0$ and separating out the $k_n$ 
term of the summation that is dependent on $D^n f^{t-1}(0)$.
The remaining $\pi(n)-1$ terms of the summation are only dependent on $D^k f^{t-1}(0)$, where $0<k<n$. Let this partial summation be written as $\sigma(n)$ with $\sigma(0)=0$ and $\sigma(1) = 1$.
Rewriting the $\pi(n)-1$ terms of the summation as $\sigma(n)$ will help in writing a proof by general induction. For $n>1$,
\begin{equation}
D^n f^t(0)=\sigma(n) + D'f(0) D^n f^{t-1}(0) 
\label{eq:Linear Equation}
\end{equation}
Theorem: The Taylor series of an iterated holomorphic function $f^t(x)$ can be constructed given a fixed point where $t \in \mathbb{N}$.
Proof. Assume the given fixed point is at zero. The Taylor series of  $f^t(x)$ can be constructed for some positive value of $R$ where $0 < |x| < R$ 
 if and only if $D^n f^t(0)$ can be constructed for every $n \geq 0$. 
 prove by strong induction. 
Basis Steps:
Case $n=0$. By definition $D^0 f^t(0) = 0$, so $D^0 f^t(0)$ can be constructed.
Case $n=1$. Let $D^1 f^t(0) = D'f(0)^t$, so $D^1 f^t(0)$ can be constructed.
Induction Step:
Case $n=k$. Assume that $D^k f^t(0)$ can be constructed for all $k$ where $0 \leq k < n$. 
Using Eq. Dynamical Recurrance Equation, $D^k f^t(0)=\sigma(k) + D'f(0) D^k f^{t-1}(0)$. The function $\sigma(k)$ in only dependent on $D^0 f(0), \ldots, D^k f(0)$ and $D^k f^t(0), \ldots, D^{(k-1)} f^t(0)$. By the strong induction hypothesis, $\sigma(k)$ can be constructed. Therefore Eq. Dynamical Recurrance Equation can be reduced to a geometrical progression based on $D'f(0)$ that can be represented by a summation. 
\begin{eqnarray}
D^k f^t(0) = \sum_{j=0}^{k-1} \sigma(k) D'f(0)^j
\end{eqnarray}
This completes the induction step that $D^n f^t(0)$ can be constructed for all whole numbers $n$. 
The Taylors series for $f^t(x)$ is
\begin{eqnarray}
f^t(x) = \sum_{n=0}^\infty \sum_{j=0}^{n-1} \sigma(n) D'f(0)^j x^n
\label{eq:Dynamical Equation}
\end{eqnarray}
$\blacksquare$
