You can easily define transformations $T$ such that $T(f(x)) = T(g(x))$ holds for some $f(x) \ne g(x)$. As you point out, a trivial example of $T$ is the $0$ transformation, i.e. $T(f(x)) = 0$ for all functions $f(x)$, then for any two functions $f$ and $g$, $T(f(x)) = T(g(x))$. For a less trivial example, let $f(x) = x$, $g(x) = x + 1$, and let $T$ be differentiation. Then $f(x) \ne g(x)$ almost everywhere, but $T(f(x)) = T(g(x)) = 1$ for all $x$.
If you mean for $T$ to be a linear transformation, then what you are interested in is the nullspace of $T$. In particular, if $f(x) - g(x)$ is in the nullspace of $T$, then we will have $T(f(x) - g(x)) = 0$, which implies $T(f(x)) = T(g(x))$. It is fairly common for linear transformations to have a non-trivial nullspace, for example differentiation, like I noted above.
If you are not just looking at linear transformations, I don't know if there's a specific terminology for cases when $T(f(x)) = T(g(x))$, but you're essentially asking about the existence of non-injective transformations (since injectivity is equivalent to $f(x) \ne g(x) \implies T(f(x)) \ne T(g(x))$), but there is no reason to expect an arbitrary transformation to be injective.