This is an annoying thing that I think most people are sloppy about describing properly, and getting it right involves being careful to maintain a clear distinction between active and passive transformations, which is almost never done in my experience.
"Matrix equivalence," properly understood, is about passive transformations (which is another name for change of coordinates / change of bases). It answers the following question:
Suppose $T : V \to W$ is a linear map between two finite-dimensional vector spaces. If $B$ is a basis of $V$ and $W$ is a basis of $C$, write $_B[T]_C$ for the matrix of $T$ expressed in these bases. As we allow both the bases $B, C$ to vary arbitrarily, what do the resulting matrices have in common?
This sounds like the question you're interested in. The answer is that the matrices $_B[T]_C$ can be any matrix that has the same dimensions and rank as $T$. We can relate this more abstract question to the concrete definition of matrix equivalence by writing down more explicitly the relationship between $_B[T]_C$ and $_{B'}[T]_{C'}$ where $B', C'$ are two arbitrary other bases; they are related by change-of-basis matrices, or more explicitly
$$_{B'}[T]_C = (_{B'}[I]_B) (_B[T]_C) (_C[I]_{C'}).$$
(This notation is slightly nonstandard but as you can see it has two very convenient properties: it naturally includes change-of-basis matrices as a special case, so those don't require a separate notation, and when multiplying matrices "the bases cancel.")
The change-of -basis matrices $_{B'} I_B$ and $_C [I]_{C'}$ can be arbitrary invertible matrices, so this is where the definition of matrix equivalence comes from. That is:
Two matrices $M, N$ are matrices of the same linear transformation $T$ with respect to some choices $B, C, B', C'$ of bases iff they are equivalent in the sense that there exist invertible matrices $P, Q$ such that $M = PNQ$, and this is true iff they have the same dimensions and rank.
Note, importantly, that the linear map $T$ has not changed in any way over the course of the argument; we aren't doing anything to the map $T$ itself, only to the matrices being used to represent it. Note also that the change-of-basis matrices are not linear transformations, in the sense that they aren't doing anything to the vector spaces $V$ and $W$. This is what makes them "passive transformations"; they are computing coordinate changes only. It is impossible to say any of this precisely without understanding clearly the definition of an abstract vector space and the way in which working with them is different from working with $\mathbb{R}^n$.
Nothing we've said contradicts the fact that there are many linear maps between any two given vector spaces, because we defined matrix equivalence as an equivalence relation on matrices, not on linear maps.
It's also true that if $T, S : V \to W$ are two linear maps with the same rank then there exist bases $B, C, B', C'$ such that $_B[T]_C = (_{B'} [S]_{C'})$, but this does not imply that $T$ and $S$ are the same linear transformation, because we're computing these matrices with respect to different bases. Actually it's a bad idea to compare matrices which are expressing linear maps with respect to different bases at all, precisely because these are not "meaningful" comparisons in the sense that they don't correspond to comparisons between linear maps. The issue is extremely similar to and in fact a generalization of comparing unital quantities measured in different units; so it's like comparing an object that's $12$ inches long to another object that's $12$ miles long. Those aren't the same length!
(There is also a second "active transformation" definition of equivalence in which we express everything in terms of linear maps and never mention bases at all. But I think it would be more confusing than helpful to discuss it here. It answers a slightly different but confusingly similar-looking question to the above question.)
