Showing that the language L={⟨M,w⟩ | M moves its head in every step while computing w} is decidable or undecidable
If $M$ does not have the option to not move its head at all, as stated in a comment to the question, then the problem is trivially decidable by the Turing machine that immediately halts and accepts.
If $M$ might not move its head while computing $T(w)$ and the problem is to decide whether it actually does so, then the problem must be undecidable as otherwise the halting problem would also be decidable.
To see this, construct a Turing machine $M'$ that simulates $M$ while always moving its head. When the simulation terminates, $M'$ keeps its head in place for one more step, and then halts. It is not hard to come up with an algorithm that builds $M'$ from $M$, thus showing that $M'$ is computable.
Then $\langle M', w \rangle \in L$ if and only if $M$ does not halt.