Does homeomorphic to itself imply the same topology?

Firstly a terminological point: the equivalence relation is called homeomorphic; a "homeomorphism" is a particular kind of function.

To answer your question: yes.

In particular, let $X$ denote a set with two or more elements. Then there exist distinct topologies $\tau,\tau' \subseteq \mathcal{P}(X)$ such that $(X,\tau)$ and $(X,\tau')$ are homeomorphic (but distinct).

Proof. Since $X$ has two or more elements, we may let $x$ and $y$ denote distinct points of $X$. Now define $\tau = \{\emptyset,\{x\},X\}$ and $\tau' = \{\emptyset,\{y\},X\}$. Observe that $\tau$ and $\tau'$ are distinct, and that they're both topologies.

Now consider the function $f : X \rightarrow X$ that permutes the points $x$ and $y$ while leaving all the other points constant. This is a homeomorphism $f : (X,\tau) \rightarrow (X,\tau'),$ thus $(X,\tau)$ is homeomorphic to $(X,\tau')$.

Well... yes. In a bit more generality that has been done so far, suppose $\mathcal{O}$ is a topology on a set $X$, and $f : X \to X$ is a bijection, then $$\mathcal{O}_f := \{ f [ U ] = \{ f(x) : x \in U \} : U \in \mathcal{O} \}$$ is also a topology on $X$. Usually this topology will differ from the original topology (but not always: what if $\mathcal{O}$ is the discrete topology? or if $f$ is the identity function?), but the mapping $f$ will be a homeomorphism from $\langle X , \mathcal{O} \rangle$ onto $\langle X , \mathcal{O}_f \rangle$.

Do you want the homeomorphism to be the identity? In that case, it will follow that the topologies must necessarily agree.

If not, consider $X = \{1,2\}$ with the two (different) topologies $\mathcal{T}_1 = \{\emptyset,\{1\},X\}$ and $\mathcal{T}_2 = \{\emptyset,\{2\},X\}$. The identity is not a homeomorphism in this case but the map interchanging $1$ and $2$ is.