kernel of $f$ and $f \circ g$ which are automorphisms of a vector space While studying image and kernel section of linear algebra, I met this problem:

Let $E$ be a vector space. Let $f$ and $g$ be two automorphisms on $E$  s.t. $f\circ g$ is the identity mapping. Show that $ker(f)$ = $ker(g\circ f)$.

I think (and I think I proved it) because $(f \circ g)$ is the identity mapping, $(g \circ f)$ is also the identity mapping of the vector space $E$. Then, based on my thought that $(g \circ f)$ is the identity mapping, $ker(g \circ f) = 0_E$, because only $0_E$ can satisfy $(g \circ f)(x) = 0_E$. However, I cannot say that $ker(f)$ is also $0_E$. $f$ is an automorphism on $E$, so there will be only one vector in $ker(f)$, but is it $0_E$? How can I say that $ker(f) = ker(g \circ f)$ satisfies?
 A: In this case, the fact that $f \circ g = \text{id}_E$ is irrelevant. Since $f$ and $g$ are automorphisms, $g \circ f$, and $f \circ g$ are also automorphisms. Hence,
\begin{align}
\ker(f) = \ker(f \circ g) = \ker(g \circ f) =\{0\}.
\end{align}
(Simply because each of these maps is injective, so their kernel is trivial.)
More interesting would be to show that for any vector space $E$, if $f:E \to E$ is linear, and $g$ is a linear automorphism of $E$ then $\ker(f) = \ker(g \circ f)$.
A: $x\in kerf$ iff $f(x)=0$ iff $g(f(x))=0$ iff $x\in ker g\circ f.$
A: Both $f:E \to E$ and $g:E \to E$ are bijective, since they're automorphisms.
In particular, since $f:E \to E$ is injective, let $x \in ker(f)$. So:
$$f(x) = 0 = f(0)$$
$$\implies x = 0$$
So, if a given linear map is injective, that means that its kernel contains only the additive identity of the vector space. You can also prove that the converse holds as well. 
Since $g \circ f$ is bijective, it is injective. So, $ker(g \circ f) = \{0\}$. Hence, $ker(f) = ker(g \circ f)$. 
Unless I've completely misinterpreted your question, the above should be fine by way of a proof. 
A: Note that $ker(f)=ker(g\circ f)$ simply because $g$ is injective 
Let's show it by double inclusion. 


*

*Let $x\in ker(f)$. Then $f(x)=0$. Since $g$ is linear, we have $g\circ f(x)=g(f(x))=g(0)=0$, so $x\in ker(g\circ f)$.

*Conversely, assume that $x\in ker(g\circ f)$. Then $g\circ f(x)=0$, so $g(f(x))=0$. This means that $f(x)\in ker(g)$. But $g$ is an automorphism, so it is injective and $ker(g)=\{0\}$. Therefore, $f(x)=0$, which means that $x\in ker(f)$.
By double inclusion, we have $ker(f)=ker(g\circ f)$.
