Die fixed so it can't roll the same number twice in a row, using markov chains?

Collapse the six states into the two states you actually care about: "$5$" and "not-$5$". There is a well-defined transition probability from each of these to the other, so we can now work with a $2 \times 2$ matrix instead of a $6 \times 6$ matrix.

This is a very common trick with Markov chains. The hard part is making sure that you don't lose any information by grouping the states together. (Here, if we had a different probability of going $1 \to 5$ than of going $2 \to 5$, this wouldn't work.)

Excellent suggestion by @Misha. Since $6$ is a small number, you may want to numerically explore this problem by computing the powers of the transition probability matrix.

$$ P = \begin{bmatrix} 0 & 1/5 & 1/5 & 1/5 & 1/5 & 1/5 \\ 1/5 & 0 &1/5 & 1/5 & 1/5 & 1/5 \\ 1/5 & 1/5 & 0 &1/5 & 1/5 & 1/5 \\ 1/5 & 1/5 & 1/5 & 0 & 1/5 & 1/5 \\ 1/5 & 1/5 & 1/5 & 1/5 & 0 & 1/5 \\ 1/5 & 1/5 & 1/5 & 1/5 & 1/5 & 0 \end{bmatrix} \enspace.$$

You'll see that $P^k$ quickly converges to a matrix where all coefficients are $1/6$. In fact, $P^5$ is already quite close. Of course, you can also repeat the experiment with the reduced chain with two states.

You can also find the steady-state probabilities (they are well-defined in this case) by solving:

$$ \begin{bmatrix} -1 & 1/5 & 1/5 & 1/5 & 1/5 & 1/5 \\ 1/5 & -1 &1/5 & 1/5 & 1/5 & 1/5 \\ 1/5 & 1/5 & -1 &1/5 & 1/5 & 1/5 \\ 1/5 & 1/5 & 1/5 & -1 & 1/5 & 1/5 \\ 1/5 & 1/5 & 1/5 & 1/5 & -1 & 1/5 \\ 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} p_1 \\ p_2 \\ p_3 \\ p_4 \\ p_5 \\ p_6 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \enspace. $$

It comes as no surprise that the solution is $[1/6, 1/6, 1/6, 1/6, 1/6, 1/6]^T$. Once again, the smart way to solve this problem is the one suggested by Misha, but sometimes numerical explorations reinforce our grasp of a concept.