Multiplier making a stop-anytime die game fair

You start with $1. Each time you roll a die and it is not a $1$, your stake is multiplied by $\alpha$; if it is a $1$, you lose everything. You may quit whenever you like and keep your current stake. For what $\alpha$ is your expected payout the same no matter how many rolls you decide to make?

$\alpha=6/5$. After committing to $n$ rolls the expected payout is $\left(\tfrac56\alpha\right)^n$. This is constant (equal to the starting $1) for all $n$ exactly when $\tfrac56\alpha=1$.