= Solution
{c}
\b[Yes.] Take $X=S^1$, the <circle>, and $A=\{1\}\subset S^1$. The <constant function> $r(z)=1$ is a <continuous map> and satisfies $ri=\operatorname{id}_A$, so $A$ is a <retract> of $X$.
However, their <fundamental groups> are
$$
\pi_1(S^1,1)\cong\mathbb Z,\qquad \pi_1(\{1\},1)=1.
$$
A <homotopy equivalence> between <path-connected spaces> induces a <group isomorphism> of their <fundamental groups>, allowing the usual change of <base point>. The <infinite cyclic group> is not the <trivial group>. Therefore \b[a <retract> need not have the <homotopy type> of its ambient space].
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