Why is Graph Isomorphism downward self reducible?
This answer shows how to prove an easier variation, the case of (undirected loopless) multigraph isomorphism.
The problem of multigraph isomorphism is downward self-reducible
The simple idea is to merge two vertices in a way that distinguish those two vertices and edges involving them.

Let $G$ be a multigraph. If the sum of degrees of two vertices reaches the maximum, we will call an ordered pair of those two vertices a degree-maximum pair. For a degree-maximum pair $(u,v)$, construct $G_{u,v}$ as follows.

Its vertices are  all vertices of $G$ except $u$ and $v$, together with a new vertex $u_v$ which is understood as $u$ merged by $v$. Note that $G_{u,v}$ has one less vertice than $G$.

Its edges consists of two parts.

the edges of $G$ whose endpoints are neither $u$ nor $v$ with the same multiplicity. Let $m$ be the maximum multiplicity of these edges.
for all vertex $w$ of $G$ that is neither $u$ nor $v$, edge $\{w, u_v\}$ with multiplicity that is the sum of the multiplicity of $\{w,u\}$ in $G$ and the product of $m+1$ and the multiplicity of $\{w,v\}$ in $G$.



It is easy to see that if we are given a multigraph $G'$ that is isomorphic to $G_{u,v}$ for some multigraph $G$, we can reconstruct $G$ from $G'$ up to isomorphism.

Suppose we are given two graphs of $n$ vertices, $G$ and $H$. Use the oracle to compare $G_{u,v}$ with $H_{x,y}$ for all tuples $(u,v,x,y)$, where $(u,v)$ is a degree-maximum pair in $G$ and $(x,y)$ is a degree-maximum pair in $H$. Once the oracle return yes, we conclude $G$ and $H$ are isomorphic; otherwise the oracle always return no, in which case we conclude $G$ and $H$ are not isomorphic. It take polynomial time to construct $G_{u,v}$ from $G$ or $H_{x,y}$ from $H$. There are less than $n^2$ tuples. So the whole algorithm takes polynomial time. QED.

I am yet to find a proof that shows the problem of (non-multi)graph isomorphism is downward self-reducible.