Let $M$ be a usual logspace Turing machine with work tape alphabet $\{Blank,0,1\}$ and $T=|M|$. Now, I define a different machine, let's call it the $T$-guess logspace machine $M^T$ of $M$. This machine is basically any logspace machine $M$ equipped with an oracle tape. $M^T$ simulates $M$. At every $T$'th time step of the simulation of $M$ the machine $M^T$ goes into special states such that in these states it nondeterministically guesses O(loglogn) bits one after the other and writes them on the oracle tape and at the end it goes back into the state from which it came again and thus submits its guess $g$ to the oracle. The oracle now simply flips the $g$'th bit ($Blank$ won't be flipped) of the work tape of $M^T$.
Now the problem.
Given: Logspace Turing machine $M$ in some encoding and $n$ in binary.
Question: Can the $T$-guess machine of $M$ guess exactly $n$ times such that $M^T$'s worktape after guess $n$ equals $\epsilon=Blank^*$?
And the question: Is there a simple argument for why this problem might be in $NL$? What is its complexity?
Note that with the machine $M^T$ in your hands you can easily guess your way through to $\epsilon$ in $n$ guesses if there exists a way to do that. So the problem is in $NL$ if $n$ were given in unary. I don't know if it is in $RL$ or $L$ though (with unary n).
EDIT: I noticed that it is unlikely to be in $NL$ because when we look at the configuration graph we are essentially looking for a path of length $n$ in a directed graph. If the graph was undirected or its adjency matrix diagonizable the problem would be trivially in $P$.
