If $T$ is an isomorphism acting on a separable Banach space $X$, can we find a countable, dense, linearly independent set $D\subset X$ such that $T(D)=D$? If $X$ is finite dimensional, then the answer is no, since $D$ would need to be finite, thus not dense. So the question is relevant only when $X$ is infinite dimensional.
The question is a follow-up question to the one below.
Every isomorphism on a separable Banach space has a completely invariant dense subset