If I've interpreted the rules correctly, then this should be the solution:
Explanation
We start off by
copying over the given digits to their rotationally symmetric cell, as well as copying the thermometers for ease of solving:
Next, we
can place the 1s. Using the knight move rule, we can deduce that R4C3 must be a 1. Now, either R1C2 or R1C4 is a 1, but if we make R1C2 a 1, we find that R2C5 would have to be a 1 as well, which is not possible since a 1 cannot occur in the middle of a thermometer. Thus, R1C4 is a 1, which then forces R2C7 to be a 1:
Now, we can note that
R5C3 has to be at least 5, since the start of the thermometer it is a part of has to be at least 2. As the ratio between R5C2 and R5C3 has to be 2, we conclude that R5C3 is either 6 or 8, and R5C2 is either 3 or 4. For ease of explanation, we'll call a cell with either 6 or 8 "blue" and a cell with either 3 or 4 "green." By the knight's move rule, we can deduce, in order, than R3C1 is blue; then, R3C4 is green; and finally, R3C3 is a 5:
Building off of this, we can now determine
the location of the remainder of the 5s. We can see that either R2C2 or R2C4 is a 5, but R2C2 being a 5 would violate the thermometer rule, so R2C4 has to be the 5, which forces R1C7 as a 5. This in turn means that R8C4 has to be blue (since the other candidate R8C2 is taken by a 5), so R7C7 has to be blue as well. Finally, we determine that due to the thermometer, R4C5 is either a 2 or a 3, and so is R3C2 - we will call that condition "red" from now on.
The next step involves a bit of "what if?" reasoning - if there's a more straightforward logical way to proceed, then let me know:
Notice that R8C1 and R8C3 are green and red, in some order. In particular, R8C3 is one of 2, 3, or 4. Now, suppose that R6C3 is a 6. This would force R8C3 to be either a 2 or a 3, since there are two numbers in between the ends of that thermometer - hence, it is red. This means R8C1 is green; in fact, R8C1 would be a 3, since R5C2 has to be a 3 by our ratio condition. Thus, the numbers in between R8C3 and 6 on that thermometer have to be 4 and 5. However, we cannot use a 5 here as we've already placed one in each row, column, and box. This gives us a contradiction, and since the only assumption we made here was that R6C3 was a 6, we conclude that R6C3 has to be an 8 instead. This means that R6C2 is a 4, making all blue cells 8 and all green cells 4:
We're at the home stretch now! Next, we see that
R8C3 has to be a 2 - if it were a 4, then the thermometer containing it would have to go "4, 6, 7, 8," and the thermometer overlapping it cannot be completed at the end; if it were a 3, then R8C1 would be a 4, and again the thermometer would have to go "3, 6, 7, 8." Thus, R8C3 is a 2, R8C1 is a 4, and all red cells are 2s. We can fill out the grid using these deductions:
And now, we can finish it off pretty simply:
R7C4, R6C4, and R6C5 have to be 3, 6, and 7 respectively. This forces R2C8 to be a 6 and R3C7 to be a 7, completing that thermometer. The rest of the grid can be completed straightforwardly, using sudoku and knight move rules as necessary to get our final solution:
