The problem is solved if there is a way to change the orientation of any specified single key, without changing any of the others. This is equivalent to finding a way to switch the chosen key an odd number of times, while switching all other keys an even number of times.
This can be done by switching all keys on the same row and column of the chosen key (including the chosen key). To see why, choose a key $K$. If we switch it and all of its sisters that share the same row and column, $K$ will be switched 7 times. Now, examine the other 15 keys in the lock. There are two cases: either the key is a sister of $K$, or not.
Suppose that $L$ is a sister of $K$, say, sharing a row with $K$. Then $L$ will be switched 4 times. For the other case, suppose $M$ is not a sister of $K$.
Then $M$ will be switched twice, because among the 6 sisters of $K$ which are turned, exactly two of them share a row or column with $M$. Consequently, of all the keys in the lock, only $K$ is switched an odd number of times.
All other keys are switched either 2 or 4 times, leaving their orientation unchanged. Thus we will be able to open the lock by selecting each horizontal key one-by-one, and turning it and all of its sister keys. $\square$