The image of a Fredholm operator under an isomorphism : did the index still the same?

Question

Let $X$, $Y$ be infinite Hilbert spaces, $T:X\to Y$ a bounded linear invertible operator and $A:X\to X$ a bounded linear operator.
Consider then the following application:
$$ \mathcal{F}: \mathscr{L}(X,X)\to \mathscr{L}(Y,Y); B\mapsto \mathcal{F}(B):=T\circ B\circ T^{-1} $$ $\mathcal{F}$ is an isomorphism (bounded, linear and invertible).
If $A$ is a Fredholm operator, then one can show using the Atkinson theorem that $\mathcal{F}(A)$ is also a Fredholm operator. But, in such a case, did we have that $A$ and $\mathcal{F}(A)$ have the same index ?
I know that the composition of two Fredholm operators $u$ and $v$ is a Fredholm operator and the index of the composition is the sum of index of $u$ and $v$. Also, the index of an invertible operator is 0. So, can we deduce that $\mathcal{F}(A) $ ans $A$ have the same index ?