How many unique "$\phi$-nary" expansions are there for $1$?

Note that from $\phi^{-1} + \phi^{-2} = 1$ it immediately follows that $\phi^{n-1} + \phi^{n-2} = \phi^n$. It follows that every valid terminating expansion that ends in 1 can be extended by replacing the final 1 by 011.

This brought you from 1 to 0.11, and from there to 0.1011, and could be repeated indefinitely.

In the limit it gives you the infinite expansion you found: $0.101010\ldots 1010\ldots$.

There are countably infinitely many finite expansions. For starting with $1$, we can replace the terminating $1$ in the $n$th phi-nimal place by $011$ in the $n$th, $n+1$th, and $n+2$th places respectively.

Now suppose given an infinite binary sequence $b$ such that $\sum b_n \phi^{-n} = 1$. Consider the following possibilities:

1. $b_0 = 1$. Then $b$ is a single $1$ followed by infinite zeroes.

2. $b_0 = 0$ and $b_1 = 1$. Then we have $\sum b_{n + 2} \phi^{-n} = 1$.

3. $b_0 = 0$ and $b_1 = 0$. Then we have $\sum b_{n + 2} \phi^{-n} \leq \frac{1}{1 - \phi^{-1}} = \phi^2$, and equality can only hold when every $b_i$ for $i \geq 2$ is 1.

Thus, it is apparent that either

1. $b$ is the alternating sequence $0, 1, 0, 1, ...$
2. $b$ begins with a prefix of the sequence $0, 1, ...$ but eventually terminates with a $1$ in an evenly indexed position or
3. $b$ begins with a prefix of the alternating sequence $0, 1, ...$ but eventually has a $0$ in an odd-indexed position, followed by an endless sequence of $1$s

So the set of all $\phi$-nary representations of $1$ is countable.