If a linear transformation has domain = range = codomain , is its kernel trivial? Suppose that a linear transformation $L$ has the same domain, range, and codomain. Then $\dim(\text{domain}) = \dim(\text{range}) + \dim(\text{kernel})$ implies that $\dim(\text{domain}) = \dim(\text{range})$.

Can we conclude from here that  $\ker L$ = $\{0\}$?
It is quite important to say whether you work in a finite dimensional space. (In this case what you wrote is true. See, for example, Show that if $W$ is a finite dimensional $K$-vector space any linear surjective map $f:W\to W$ is bijective.)

If you allow infinite dimensional spaces, this is not necessarily true. 

Take the space $V=\mathbb R^{\mathbb N}$ of all real sequences. Define $L\colon V\to V$ as
$$L \colon (x_1,x_2,x_3,\dots) \mapsto (x_2,x_3,x_4,\dots).$$
Then the range is the whole space $V$. But the map is not injective. For example, the sequence $(1,0,0,\dots)$ is in the kernel.
Yes. It means the transformation is bijective.