The $n$th mathematician needs to store the position of the previously imprisoned mathematicians respecting the order. He knows in which order the other mathematicians have been selected, so he just needs to store the cell numbers in order. The total number of posible variations is $7!/(7-n)!$. For $n=5$ we have $2520$ variations, so we have enough clock positions to encode all variations.

The key is that at this point everyone knows who the first mathematician is and where he was improsoned, so they only have to know where the remaining 6 mathematicians were imprisoned in the remaining 6 cells. This means that there are $6!=720$ possible permutations that need to be encoded. This method will not reveal the order in which they have been imprisoned, only what mathematician has been assigned to each cell. For example, let us assume that Cid was the first mathematician and he was assigned cell #2. His name and cell number need to be removed from the name and cell number lists respectively, so that cell #3 now represents the 2nd cell and so on (the same can be done for the name list).

The $n$th mathematician needs to store the position of the previously imprisoned mathematicians respecting the order. They know the order in which the other mathematicians have been selected, so they just needs to store the cell numbers in order. The total number of posible variations is $7!/(7-n)!$. For $n=5$ we have $2520$ variations, so we have enough clock positions to encode all variations.

The key is that at this point everyone knows who the first mathematician is and where he was improsoned, so they only have to know where the remaining 6 mathematicians were imprisoned in the remaining 6 cells. This means that there are $6!=720$ possible permutations that need to be encoded. This method will not reveal the order in which they have been imprisoned, only what mathematician has been assigned to each cell. For example, let us assume that Cid was the first mathematician and he was assigned cell #2. His name and cell number need to be removed from the name and cell number lists respectively, so that cell #3 now represents the 2nd cell and so on (the same can be done for the name list).

For example, in this particular case, the 1A 3B 4D 5E 6F 7G configuration (2C is implied) would be represented by number 0 (00:00), the 1A 3B 4D 5E 6G 7F configuration by number 1 (00:01), 1A 3B 4D 5G 6E 7F by number 2 (00:02), 1A 3B 4D 5G 6F 7E by number 3 (00:03), and so on.

In the following I describe a method for the mathematicians to pass the information to one another using the clock so that they can all determine in which cell each mathematician is imprisoned. My method assumes that the two hands of the clock can be arbitrarily rotated. In conventional clocks, the hour and minute hands are mechanically coupled through gears (the position of the hour hand at a given hour is determined by the position of the minute hand) so that the clock only has $12 \times 60=720$ positions. If the two hands are uncoupled, you can use the $60 \times 60=3600$ positions of the clock to send information. I will use the following notation to express the position of the hands: hh:mm, where the two digit numbers hh and mm range from 0 to 59. This is equivalent to using a sexagesimal numeral system, so that hh:mm represents the hh*60+mm number. For example, if the hour hand points over the 33 minute mark and the minute hand over the 15 minute mark the position of the clock is 33:15, which represents the 1995 number in the decimal system.

The OP mentions that the mathematicians are initially in the same room, so they know who has been selected before them and in which order.

The mathematicians need to use the clock to store where the previously selected mathematicians and themselves have been imprisoned. The first mathematician only needs to store his position, there are 7 posible options so he only needs to cover the numbers from 0 to 6 (00:00 to 00:06).

The first 5 mathematicians should follow the same procedure:

The $n$th mathematician needs to store the position of the previously imprisoned mathematicians respecting the order. He knows in which order the other mathematicians have been selected, so he just needs to store the cell numbers in order. The total number of posible variations is $7!/(7-n)!$. For $n=5$ we have $2520$ variations, so we have enough clock positions to encode all variations.

When the 6th mathematician enters the transfer room, he will know where each of the previous mathematicians have been imprisoned, plus where he will be imprisoned. This means that they know that the 7th mathematician will be imprisoned in the remaining cell. The same thing can be said about the 7th mathematician, once he sees the clock prepared by the 5th mathematician he will know the complete solution to the problem.

The 7th mathematician needs to transfer this knowledge to the rest of mathematicians, specially to the first one, who has the least information so far (he only knows that he was imprisoned first and their cell number).

The key is that at this point everyone knows who the first mathematician is and where he was improsoned, so they only have to know where the remaining 6 mathematicians were stored in remaining 6 cells. This means that there are $6!=720$ possible permutations that need to be encoded. This method will not reveal the order in which they have been imprisoned, only what mathematician has been assigned to each cell. For example, let us assume that Cid was the first mathematician and he was assigned cell #2. His name and cell number need to be removed from the name and cell number lists respectively, so that cell #3 now represents the 2nd cell and so on (the same can be done for the name list).

This way, all mathematicians will know what mathematician has been assigned to each cell when they walk through the transfer room in their way back to the king.

In the following I describe a method for the mathematicians to pass the information to one another using the clock so that they can all determine in which cell each mathematician is imprisoned. My method assumes that the two hands of the clock can be arbitrarily rotated. In conventional clocks, the hour and minute hands are mechanically coupled through gears (the position of the hour hand at a given hour is determined by the position of the minute hand) so that the clock only has $12 \times 60=720$ positions. If the two hands are uncoupled, you can use the $60 \times 60=3600$ positions of the clock to send information. I will use the following notation to express the position of the hands: hh:mm, where the two digit numbers hh and mm range from 0 to 59. This is equivalent to using a sexagesimal numeral system, so that hh:mm represents the hh*60+mm number. For example, if the hour hand points over the 33 minute mark and the minute hand over the 15 minute mark the position of the clock is 33:15, which represents the 1995 number in the decimal system.

The OP mentions that the mathematicians are initially in the same room, so they know who has been selected before them and in which order.

The mathematicians need to use the clock to store where the previously selected mathematicians and themselves have been imprisoned. The first mathematician only needs to store his position, there are 7 posible options so he only needs to cover the numbers from 0 to 6 (00:00 to 00:06).

The first 5 mathematicians should follow the same procedure:

The $n$th mathematician needs to store the position of the previously imprisoned mathematicians respecting the order. He knows in which order the other mathematicians have been selected, so he just needs to store the cell numbers in order. The total number of posible variations is $7!/(7-n)!$. For $n=5$ we have $2520$ variations, so we have enough clock positions to encode all variations.

When the 6th mathematician enters the transfer room, he will know where each of the previous mathematicians have been imprisoned, plus where he will be imprisoned. This means that they know that the 7th mathematician will be imprisoned in the remaining cell. The same thing can be said about the 7th mathematician, once he sees the clock prepared by the 5th mathematician he will know the complete solution to the problem.

The 7th mathematician needs to transfer this knowledge to the rest of mathematicians, specially to the first one, who has the least information so far (he only knows that he was imprisoned first and their cell number).

The key is that at this point everyone knows who the first mathematician is and where he was improsoned, so they only have to know where the remaining 6 mathematicians were imprisoned in the remaining 6 cells. This means that there are $6!=720$ possible permutations that need to be encoded. This method will not reveal the order in which they have been imprisoned, only what mathematician has been assigned to each cell. For example, let us assume that Cid was the first mathematician and he was assigned cell #2. His name and cell number need to be removed from the name and cell number lists respectively, so that cell #3 now represents the 2nd cell and so on (the same can be done for the name list).

This way, all mathematicians will know what mathematician has been assigned to each cell when they walk through the transfer room in their way back to the king.