Solution:
The above equation holds for infinitely many $x$ if and only if
$$(x-243) f(3x) = 729(x-3) f(x)$$
for all $x \in \mathbb{C}$, because $(x-243) f(3x) - 729(x-3) f(x)$ is a polynomial, which has infinitely many zeroes if and only if it is identically $0$.

We now plug in different values to find various zeroes of $f$:
$$\begin{aligned} x=3 & \Longrightarrow f(9)=0 \\ x=9 & \Longrightarrow f(27)=0 \\ x=27 & \Longrightarrow f(81)=0 \\ x=81 & \Longrightarrow f(243)=0 \end{aligned}$$
We may write $f(x) = (x-243)(x-81)(x-27)(x-9) p(x)$ for some polynomial $p(x)$. We want to solve
$$\begin{aligned} & (x-243)(3x-243)(3x-81)(3x-27)(3x-9) p(3x) \\ & \quad = 729(x-3)(x-243)(x-81)(x-27)(x-9) p(x). \end{aligned}$$
Dividing out common factors from both sides, we get $p(3x) = 9 p(x)$, so the polynomial $p$ is homogeneous of degree $2$. Therefore $p(x) = a x^{2}$ and thus $f(x) = a(x-243)(x-81)(x-27)(x-9)x^{2}$, where $a \in \mathbb{C}$ is arbitrary.

Alternative solution:
As above, we wish to find polynomials $f$ such that
$$(x-243) f(3x) = 729(x-3) f(x).$$
Suppose $f$ is such a polynomial and let $Z$ be its multiset of zeroes.
Both $(x-243) f(3x)$ and $729(x-3) f(x)$ have the same multiset of zeroes, i.e. $\{243\} \cup \frac{1}{3} Z = \{3\} \cup Z$ or $\{729\} \cup Z = \{9\} \cup 3Z$. Thus
$$\begin{aligned} 9 \in Z & \Rightarrow 27 \in 3Z \\ & \Rightarrow 27 \in Z \\ & \Rightarrow 81 \in 3Z \\ & \Rightarrow 81 \in Z \\ & \Rightarrow 243 \in 3Z \\ & \Rightarrow 243 \in Z \end{aligned}$$
Let $Y$ be the unique multiset with $Z = \{9,27,81,243\} \cup Y$. This gives
$$\{9,27,81,243,729\} \cup Y = \{9,27,81,243,729\} \cup 3Y$$
so $Y \subset \mathbb{C}$ is a finite multiset, invariant under multiplication by $3$. It follows that $Y = \{0,0,\ldots,0\}$ with some multiplicity $k$. Hence
$$f(x) = a x^{k}(x-9)(x-27)(x-81)(x-243)$$
for some constant $a \in \mathbb{C}$. Plugging this back into the original equation, we see that $k=2$ and that $a \in \mathbb{C}$ can be arbitrary.