Let $A \subset X$ and let $j: A \to X$ be the inclusion map. Let $f: X \to A$ be a continuous map. We suppose there is a homotopy between $j \circ f$ and the identity map on $X$. We are to show that if $f$ is a retraction, then $j_*$ (induced homomorphism) is an isomorphism.
There seems to be a fundamental misunderstanding on my part, in this scenario, I think of fixed point(s) $a, j(a) = a$ and consider $j$ as a map $$j: (A, a) \to (X, a)$$ and think of the induced homomorphism as $$j_*: \pi_1(A, a) \to \pi_1(X,a) \\ [g] \mapsto [j \circ g]$$ while the solutions I found online consider it as a map $$j_*: [A, X] \to [X, X]$$ where $[X, Y]$ denotes the set of homotopy classes from $X$ to $Y$. While such a view does, to a certain degree make sense in this setting I must admit I fail to see why its allowed. To me it seems that we are no longer considering members of the equivalence class, but rather the equivalence classes themselves.
Maybe this solution clears it up for you:
We have that $f:X \to A$ is a retraction. So that then the restriction $f|_{A}$ is the identity map on $A$($\subset X$).
As $j: A \to X$, then the composition $f \circ j$ is the identity map on $A$. It follows that this composition induces a homomorphism $f_{*}\circ j_{*}$ which is the identity map on $\pi_{1}(A,a)$.
In particular, for any loops $[k],[l] \in \pi_{1}(A,a)$ such that $j_{*}[k]=j_{*}[l]$, then we have
$(f_{*}\circ j_{*})[k]=(f_{*} \circ j_{*})[l]$ implies $[k]=[l]$, as $f_{*}\circ j_{*}$ is the identity map on $\pi_{1}(A,a).$ It follows at once that $j_{*}$ is injective.
Note that, it is not necessarily true that $j_{*}$ is a surjection. Thusly, it's not necessarily an isomorphism.