Nothing wrong with Anthony Quas's answer, and indeed the answer is "no", but I'll just remark that you already asked essentially the same question before, and the same answer works directly, with $X = Y$. Let me recall my answer to this question of yours.

"The answer is yes [the previous question was phrased differently]. In this paper Downarowicz proves the following theorem

There exists a regular Toeplitz sequence $\omega$ such that the induced
Toeplitz flow $(\bar O(\omega), S)$ is noncoalescent, more precisely, it admits an endomorphism $\pi : \bar{O}(\omega) \to \bar{O}(\omega)$ of the first type.

Here, $\bar O(\omega)$ is the orbit closure (so a Toeplitz subshift since $\omega$ is Toeplitz), $S$ the shift map, **noncoalescent** means not injective, and **first type** means every Toeplitz point has a unique preimage, in particular some point does."

To see that this answers your question, set $X = Y$, and $\tilde Y$ the Toeplitz points of $Y$, so that every point in $\tilde Y$ has unique preimage. What we need is that the Toeplitz points are a residual set, and that their measure is $1$ in every invariant probability measure. Both facts are well-known. The first fact is mentioned e.g. on page 23 of this paper (it works for any Toeplitz subshift), and the second is Theorem 13.1. in the same reference (it works for any regular Toeplitz subshift).