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Maybe it also makes sense to provide an "abstract" answer to this. Place yourself in the category of spaces over $M$. For present purposes it does not need to be a manifold, and in fact the category we work in may be more general too. I will suppress $M$, i. e. when I say "an object $X$" I will actually mean a map $X\to M$, and when I say "a morphism $Y\to Z$" I will actually mean a fibrewise map from $Y\to M$ to $Z\to M$, etc.

Now for an object $X$, there is another one $\operatorname{Aut}(X)$, with the property that morphisms from any $Y$ to $\operatorname{Aut}(X)$ are in one-to-one correspondence with fibrewise automorphisms of $Y\times_MX\to Y$ over $Y$. This has canonical group structures on its fibres; it might be a group bundle (but not necessarily) and this group bundle might even be trivial, i. e. be isomorphic (over $M$) to the projection $M\times G\to M$ for some "usual" group $G$, and then one gets "your" situation.

More generally, for another object $X'$, there is still another one $\operatorname{Iso}(X,X')$ such that fibrewise maps from any $Y$ to $\operatorname{Iso}(X,X')$ are in one-to-one correspondence with fibrewise isomorphisms between $Y\times_MX\to Y$ and $Y\times_MX'\to Y$ over $Y$.

Now $\operatorname{Iso}(X,X')$ comes with a canonical right $\operatorname{Aut}(X)$-action and a canonical left $\operatorname{Aut}(X')$-action which commute, and would be a principal left $\operatorname{Aut}(X')$- and a principal right $\operatorname{Aut}(X)$-bundle in appropriate sense, except that it may happen not to have global support - i. e. the map $\operatorname{Iso}(X,X')\to M$ may happen not to be surjective (in fact it might well happen that $\operatorname{Iso}(X,X')$ is empty).

However if that map is surjective, then you may try to reconstruct $X'$ from $X$ together with $\operatorname{Iso}(X,X')$. More precisely it is natural to hope that in this case $X'$ is isomorphic to the (fibrewise) quotient of $\operatorname{Iso}(X,X')\times X$ by $\operatorname{Aut}(X)$; informally speaking, you take pairs $\left\langle\varphi,x\right\rangle$ where $x\in X$ and $\varphi:X\to X'$, and identify $\left\langle\varphi,\alpha(x)\right\rangle$ with $\left\langle\varphi\circ\alpha,x\right\rangle$ for $\alpha\in\operatorname{Aut}(X)$. This is the associated bundle construction.

All this works for sure when $X$ is a trivial bundle, i. e. is the projection $M\times F\to M$ for some $F$; then (in good cases) $\operatorname{Aut}(X)$ will be the trivial group bundle $M\times G\to M$. Given some structure on $F$ (such that a vector space) you might want to further switch to the subgroup consisting of structure preserving automorphisms, and to the subbundle of $\operatorname{Iso}(X,X')$ consisting of structure preserving (fibrewise) isomorphisms. You then get as one of the examples what you are after: a vector bundle is more or less the same thing as a fibrewise vector space $E\to M$ such that for a trivial one $M\times V\to M$ the object $\operatorname{Iso}(M\times V,E)$ (which can be identified with the frame bundle of $E$ once you choose a basis in $V$) has global support.

I cannot think of an appropriate reference right away but there surely are many, all this is well known at least from 1970ies on...

Maybe it also makes sense to provide an "abstract" answer to this. Place yourself in the category of spaces over $M$. For present purposes it does not need to be a manifold, and in fact the category we work in may be more general too. I will suppress $M$, i. e. when I say "an object $X$" I will actually mean a map $X\to M$, and when I say "a morphism $Y\to Z$" I will actually mean a fibrewise map from $Y\to M$ to $Z\to M$, etc.

Now for an object $X$, there is another one $\operatorname{Aut}(X)$, with the property that morphisms from any $Y$ to $\operatorname{Aut}(X)$ are in one-to-one correspondence with fibrewise automorphisms of $Y\times_MX\to Y$ over $Y$. This has canonical group structures on its fibres; it might be a group bundle (but not necessarily) and this group bundle might even be trivial, i. e. be isomorphic (over $M$) to the projection $M\times G\to M$ for some "usual" group $G$, and then one gets "your" situation.

More generally, for another object $X'$, there is still another one $\operatorname{Iso}(X,X')$ such that fibrewise maps from any $Y$ to $\operatorname{Iso}(X,X')$ are in one-to-one correspondence with fibrewise isomorphisms between $Y\times_MX\to Y$ and $Y\times_MX'\to Y$ over $Y$.

Now $\operatorname{Iso}(X,X')$ comes with a canonical right $\operatorname{Aut}(X)$-action and a canonical left $\operatorname{Aut}(X')$-action which commute, and would be a principal left $\operatorname{Aut}(X')$- and a principal right $\operatorname{Aut}(X)$-bundle in appropriate sense, except that it may happen not to have global support - i. e. the map $\operatorname{Iso}(X,X')\to M$ may happen not to be surjective (in fact it might well happen that $\operatorname{Iso}(X,X')$ is empty).

However if that map is surjective, then you may try to reconstruct $X'$ from $X$ together with $\operatorname{Iso}(X,X')$. More precisely it is natural to hope that in this case $X'$ is isomorphic to the (fibrewise) quotient of $\operatorname{Iso}(X,X')\times X$ by $\operatorname{Aut}(X)$; informally speaking, you take pairs $\left\langle\varphi,x\right\rangle$ where $x\in X$ and $\varphi:X\to X'$, and identify $\left\langle\varphi,\alpha(x)\right\rangle$ with $\left\langle\varphi\circ\alpha,x\right\rangle$ for $\alpha\in\operatorname{Aut}(X)$. This is the associated bundle construction.

All this works for sure when $X$ is a trivial bundle, i. e. is the projection $M\times F\to M$ for some $F$; then (in good cases) $\operatorname{Aut}(X)$ will be the trivial group bundle $M\times G\to M$. Given some structure on $F$ (such that a vector space) you might want to further switch to the subgroup consisting of structure preserving automorphisms, and to the subbundle of $\operatorname{Iso}(X,X')$ consisting of structure preserving (fibrewise) isomorphisms. You then get as one of the examples what you are after: a vector bundle is more or less the same thing as a fibrewise vector space $E\to M$ such that for a trivial one $M\times V\to M$ the object $\operatorname{Iso}(M\times V,E)$ has global support; the latter can then be identified with the frame bundle of $E$ once you choose a basis in $V$.

I cannot think of an appropriate reference right away but there surely are many, all this is well known at least from 1970ies on...

Maybe it also makes sense to provide an "abstract" answer to this. Place yourself in the category of spaces over $M$. For present purposes it does not need to be a manifold, and in fact the category we work in may be more general too. I will suppress $M$, i. e. when I say "an object $X$" I will actually mean a map $X\to M$, and when I say "a morphism $Y\to Z$" I will actually mean a fibrewise map from $Y\to M$ to $Z\to M$, etc.

Now for an object $X$, there is another one $\operatorname{Aut}(X)$, with the property that morphisms from any $Y$ to $\operatorname{Aut}(X)$ are in one-to-one correspondence with fibrewise automorphisms of $Y\times_MX\to Y$ over $Y$. This has canonical group structures on its fibres; it might be a group bundle (but not necessarily) and this group bundle might even be trivial, i. e. be isomorphic (over $M$) to the projection $M\times G\to M$ for some "usual" group $G$, and then one gets "your" situation.

More generally, for another object $X'$, there is still another one $\operatorname{Iso}(X,X')$ such that fibrewise maps from any $Y$ to $\operatorname{Iso}(X,X')$ are in one-to-one correspondence with fibrewise isomorphisms between $Y\times_MX\to Y$ and $Y\times_MX'\to Y$ over $Y$.

Now $\operatorname{Iso}(X,X')$ comes with a canonical right $\operatorname{Aut}(X)$-action and a canonical left $\operatorname{Aut}(X')$-action which commute, and would be a principal left $\operatorname{Aut}(X')$- and a principal right $\operatorname{Aut}(X)$-bundle in appropriate sense, except that it may happen not to have global support - i. e. the map $\operatorname{Iso}(X,X')\to M$ may happen not to be surjective (in fact it might well happen that $\operatorname{Iso}(X,X')$ is empty).

However if that map is surjective, then you may try to reconstruct $X'$ from $X$ together with $\operatorname{Iso}(X,X')$. More precisely it is natural to hope that in this case $X'$ is isomorphic to the (fibrewise) quotient of $\operatorname{Iso}(X,X')\times X$ by $\operatorname{Aut}(X)$; informally speaking, you take pairs $\left\langle\varphi,x\right\rangle$ where $x\in X$ and $\varphi:X\to X'$, and identify $\left\langle\varphi,\alpha(x)\right\rangle$ with $\left\langle\varphi\circ\alpha,x\right\rangle$ for $\alpha\in\operatorname{Aut}(X)$. This is the associated bundle construction.

All this works for sure when $X$ is a trivial bundle, i. e. is the projection $M\times F\to M$ for some $F$; then $\operatorname{Aut}(X)$ will contain as a subgroup bundle the trivial one $M\times G\to M$ (maybe even coincide with it?) and you may restrict to this. Given some structure on $F$ (such that a vector space) you might want to further switch to the (fibrewise) subgroup consisting of structure preserving automorphisms, and to the subbundle of $\operatorname{Iso}(X,X')$ consisting of structure preserving (fibrewise) isomorphisms. You then get as one of the examples what you are after: a vector bundle is more or less the same thing as a fibrewise vector space $E\to M$ such that for a trivial one $M\times V\to M$ the object $\operatorname{Iso}(M\times V,E)$ has global support.

I cannot think of an appropriate reference right away that there surely are many, all this is well known at least from 1970ies on...

Maybe it also makes sense to provide an "abstract" answer to this. Place yourself in the category of spaces over $M$. For present purposes it does not need to be a manifold, and in fact the category we work in may be more general too. I will suppress $M$, i. e. when I say "an object $X$" I will actually mean a map $X\to M$, and when I say "a morphism $Y\to Z$" I will actually mean a fibrewise map from $Y\to M$ to $Z\to M$, etc.

Now for an object $X$, there is another one $\operatorname{Aut}(X)$, with the property that morphisms from any $Y$ to $\operatorname{Aut}(X)$ are in one-to-one correspondence with fibrewise automorphisms of $Y\times_MX\to Y$ over $Y$. This has canonical group structures on its fibres; it might be a group bundle (but not necessarily) and this group bundle might even be trivial, i. e. be isomorphic (over $M$) to the projection $M\times G\to M$ for some "usual" group $G$, and then one gets "your" situation.

More generally, for another object $X'$, there is still another one $\operatorname{Iso}(X,X')$ such that fibrewise maps from any $Y$ to $\operatorname{Iso}(X,X')$ are in one-to-one correspondence with fibrewise isomorphisms between $Y\times_MX\to Y$ and $Y\times_MX'\to Y$ over $Y$.

Now $\operatorname{Iso}(X,X')$ comes with a canonical right $\operatorname{Aut}(X)$-action and a canonical left $\operatorname{Aut}(X')$-action which commute, and would be a principal left $\operatorname{Aut}(X')$- and a principal right $\operatorname{Aut}(X)$-bundle in appropriate sense, except that it may happen not to have global support - i. e. the map $\operatorname{Iso}(X,X')\to M$ may happen not to be surjective (in fact it might well happen that $\operatorname{Iso}(X,X')$ is empty).

However if that map is surjective, then you may try to reconstruct $X'$ from $X$ together with $\operatorname{Iso}(X,X')$. More precisely it is natural to hope that in this case $X'$ is isomorphic to the (fibrewise) quotient of $\operatorname{Iso}(X,X')\times X$ by $\operatorname{Aut}(X)$; informally speaking, you take pairs $\left\langle\varphi,x\right\rangle$ where $x\in X$ and $\varphi:X\to X'$, and identify $\left\langle\varphi,\alpha(x)\right\rangle$ with $\left\langle\varphi\circ\alpha,x\right\rangle$ for $\alpha\in\operatorname{Aut}(X)$. This is the associated bundle construction.

All this works for sure when $X$ is a trivial bundle, i. e. is the projection $M\times F\to M$ for some $F$; then (in good cases) $\operatorname{Aut}(X)$ will be the trivial group bundle $M\times G\to M$. Given some structure on $F$ (such that a vector space) you might want to further switch to the subgroup consisting of structure preserving automorphisms, and to the subbundle of $\operatorname{Iso}(X,X')$ consisting of structure preserving (fibrewise) isomorphisms. You then get as one of the examples what you are after: a vector bundle is more or less the same thing as a fibrewise vector space $E\to M$ such that for a trivial one $M\times V\to M$ the object $\operatorname{Iso}(M\times V,E)$ (which can be identified with the frame bundle of $E$ once you choose a basis in $V$) has global support.

I cannot think of an appropriate reference right away but there surely are many, all this is well known at least from 1970ies on...

Maybe it also makes sense to provide an "abstract" answer to this. Place yourself in the category of spaces over $M$. For present purposes it does not need to be a manifold, and in fact the category we work in may be more general too. I will suppress $M$, i. e. when I say "an object $X$" I will actually mean a map $X\to M$, and when I say "a morphism $Y\to Z$" I will actually mean a fibrewise map from $Y\to M$ to $Z\to M$, etc.

Now for an object $X$, there is another one $\operatorname{Aut}(X)$, with the property that morphisms from any $Y$ to $\operatorname{Aut}(X)$ are in one-to-one correspondence with fibrewise automorphisms of $Y\times_MX\to Y$ over $Y$. This has canonical group structures on its fibres; it might be a group bundle (but not necessarily) and this group bundle might even be trivial, i. e. be isomorphic (over $M$) to the projection $M\times G\to M$ for some "usual" group $G$, and then one gets "your" situation.

More generally, for another object $X'$, there is still another one $\operatorname{Iso}(X,X')$ such that fibrewise maps from any $Y$ to $\operatorname{Iso}(X,X')$ are in one-to-one correspondence with fibrewise isomorphisms between $Y\times_MX\to Y$ and $Y\times_MX'\to Y$ over $Y$.

Now $\operatorname{Iso}(X,X')$ comes with a canonical right $\operatorname{Aut}(X)$-action and a canonical left $\operatorname{Aut}(X')$-action which commute, and would be a principal left $\operatorname{Aut}(X')$- and a principal right $\operatorname{Aut}(X)$-bundle in appropriate sense, except that it may happen not to have global support - i. e. the map $\operatorname{Iso}(X,X')\to M$ may happen not to be surjective (in fact it might well happen that $\operatorname{Iso}(X,X')$ is empty).

However if that map is surjective, then you may try to reconstruct $X'$ from $X$ together with $\operatorname{Iso}(X,X')$. More precisely it is natural to hope that in this case $X'$ is isomorphic to the (fibrewise) quotient of $\operatorname{Iso}(X,X')\times X$ by $\operatorname{Aut}(X)$; informally speaking, you take pairs $\left\langle\varphi,x\right\rangle$ where $x\in X$ and $\varphi:X\to X'$, and identify $\left\langle\varphi,\alpha(x)\right\rangle$ with $\left\langle\varphi\circ\alpha,x\right\rangle$ for $\alpha\in\operatorname{Aut}(X)$. This is the associated bundle construction.

All this works for sure when $X$ is a trivial bundle, i. e. is the projection $M\times F\to M$ for some $F$; then $\operatorname{Aut}(X)$ will be a trivial group bundle. Given some structure on $F$ (such that a vector space) you might want to switch to the (fibrewise) subgroup of $\operatorname{Aut}(X)$ consisting of structure preserving automorphisms, and to the (fibrewise) subbundle of $\operatorname{Iso}(X,X')$ consisting of structure preserving isomorphisms. You then get as one of the examples what you are after: a vector bundle is more or less the same thing as a fibrewise vector space $E\to M$ such that for a trivial one $M\times V\to M$ the object $\operatorname{Iso}(M\times V,E)$ has global support.

I cannot think of an appropriate reference right away that there surely are many, all this is well known at least from 1970ies on...

Maybe it also makes sense to provide an "abstract" answer to this. Place yourself in the category of spaces over $M$. For present purposes it does not need to be a manifold, and in fact the category we work in may be more general too. I will suppress $M$, i. e. when I say "an object $X$" I will actually mean a map $X\to M$, and when I say "a morphism $Y\to Z$" I will actually mean a fibrewise map from $Y\to M$ to $Z\to M$, etc.

Now for an object $X$, there is another one $\operatorname{Aut}(X)$, with the property that morphisms from any $Y$ to $\operatorname{Aut}(X)$ are in one-to-one correspondence with fibrewise automorphisms of $Y\times_MX\to Y$ over $Y$. This has canonical group structures on its fibres; it might be a group bundle (but not necessarily) and this group bundle might even be trivial, i. e. be isomorphic (over $M$) to the projection $M\times G\to M$ for some "usual" group $G$, and then one gets "your" situation.

More generally, for another object $X'$, there is still another one $\operatorname{Iso}(X,X')$ such that fibrewise maps from any $Y$ to $\operatorname{Iso}(X,X')$ are in one-to-one correspondence with fibrewise isomorphisms between $Y\times_MX\to Y$ and $Y\times_MX'\to Y$ over $Y$.

Now $\operatorname{Iso}(X,X')$ comes with a canonical right $\operatorname{Aut}(X)$-action and a canonical left $\operatorname{Aut}(X')$-action which commute, and would be a principal left $\operatorname{Aut}(X')$- and a principal right $\operatorname{Aut}(X)$-bundle in appropriate sense, except that it may happen not to have global support - i. e. the map $\operatorname{Iso}(X,X')\to M$ may happen not to be surjective (in fact it might well happen that $\operatorname{Iso}(X,X')$ is empty).

However if that map is surjective, then you may try to reconstruct $X'$ from $X$ together with $\operatorname{Iso}(X,X')$. More precisely it is natural to hope that in this case $X'$ is isomorphic to the (fibrewise) quotient of $\operatorname{Iso}(X,X')\times X$ by $\operatorname{Aut}(X)$; informally speaking, you take pairs $\left\langle\varphi,x\right\rangle$ where $x\in X$ and $\varphi:X\to X'$, and identify $\left\langle\varphi,\alpha(x)\right\rangle$ with $\left\langle\varphi\circ\alpha,x\right\rangle$ for $\alpha\in\operatorname{Aut}(X)$. This is the associated bundle construction.

All this works for sure when $X$ is a trivial bundle, i. e. is the projection $M\times F\to M$ for some $F$; then $\operatorname{Aut}(X)$ will contain as a subgroup bundle the trivial one $M\times G\to M$ (maybe even coincide with it?) and you may restrict to this. Given some structure on $F$ (such that a vector space) you might want to further switch to the (fibrewise) subgroup consisting of structure preserving automorphisms, and to the subbundle of $\operatorname{Iso}(X,X')$ consisting of structure preserving (fibrewise) isomorphisms. You then get as one of the examples what you are after: a vector bundle is more or less the same thing as a fibrewise vector space $E\to M$ such that for a trivial one $M\times V\to M$ the object $\operatorname{Iso}(M\times V,E)$ has global support.

I cannot think of an appropriate reference right away that there surely are many, all this is well known at least from 1970ies on...