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}

$\blacksquare$