Does this theorem/collection of theorems have a name? Or is it just seen as obvious?

This is an artifact of category theory. The actual theorem it describes is rather trivial - the hard part is choosing the right definitions.

In particular, a category can be defined by a collection of objects with maps between them called morphisms. For instance, there is a category of sets where the objects are sets, and a morphism $X\rightarrow Y$ is simply a function $f:X\rightarrow Y$. There is a category of groups, whose objects are groups $(X,\cdot)$ and a map $(X,\cdot)\rightarrow (Y,*)$ is a homomorphism between these groups. You can similarly define a category of operations of a given arity (or, in universal algebra, of a given signature - which encompasses objects like rings).

More or less, a category is supposed to have a notion of composition - meaning that if I take a map $f:X\rightarrow Y$ and a map $g:Y\rightarrow Z$, then there is a map $g\circ f:X\rightarrow Z$. Functions and homomorphisms are closed under composition, so our examples are good. One requires that composition be associative and that, for every object $X$, there is an identity map $\operatorname{id}_X:X\rightarrow X$ which is an identity for composition.

Then, one defines an isomorphism as follows:

An isomorphism $f:X\rightarrow Y$ is a morphism such that there exists an inverse $g:Y\rightarrow X$. That is, a map such that $f\circ g = \operatorname{id}_Y$ and $g\circ f=\operatorname{id}_X$. This $g$ is known as $f^{-1}$.

So, in the category of sets, the isomorphisms are the bijections. In the category of groups, the isomorphisms are exactly what is usually called an isomorphism - and your extended notion of isomorphism also corresponds to a categorical one in the right category.

We also have the following definition:

An automorphism of an object $X$ is an isomorphism $f:X\rightarrow X$.

Then, your claim is essentially the following:

If $f:X\rightarrow Y$ and $g:X\rightarrow Y$ are an isomorphisms, then there is a $\sigma\in \operatorname{Aut}(X)$ such that $g=f\circ \sigma$ and a $\sigma'\in\operatorname{Aut}(Y)$ such that $g=\sigma'\circ f$.

This is easy to show after proving a few useful lemmas from the two given definitions: First, show that the composition of isomorphisms are isomorphisms. Second, show that the inverse of an isomorphism is an isomorphism. These both follow quickly from the axioms above. Then, let $\sigma = g\circ f^{-1}$ and $\sigma'=f^{-1}\circ g$. These are isomorphisms by the two lemmas I suggest, but are also maps $X\rightarrow X$ and $Y\rightarrow Y$ - thus automorphisms - by the properties of composition.