If G has trivial center, what does Aut(G) ≈ Inn(G) imply?

If H has trivial center. I know that if Z(G) is trivial, then Aut(G) has trivial center, but I need help understanding what is happening when G has a trivial center and Aut(G) ≈ Inn(G).

I know that when Z(G) is trivial, then G ≈ Inn(G) but when does Aut(G) ≈ Inn(G) imply?

Re: If G has trivial center, what does Aut(G) ≈ Inn(G) imply?

Also, just remembered that [G:Z(G)] = #G (order of G) when Z(G) is trivial, and if the index of the centralizer is #G, we have at most #G ways to conjugate any element of G.

Since Inn(G) ≈ G/Z(G) and Z(G) is trivial, we have Inn(G) ≈ G. Say ɸ is the homomorphism that maps G to Inn(G). Since we know this homomorphism is an isomorphism (bijective), does this mean that every element of G is conjugate to all elements of G?

Feel like I'm making progress, but still not sure what Aut(G) ≈ Inn(G) is telling us.

Re: If G has trivial center, what does Aut(G) ≈ Inn(G) imply?

Not sure if I'm exactly right, but if we have Aut(G) ≈ Inn(G), then Aut(G) and Inn(G) must have the same number of elements, and since I stated in the earlier post that we found #Inn(G) = #G (So Inn(G) contains every element of G), then this means that #Aut(G) = #G. Since #Aut(G) = #Inn(G) = #G.

If the order of the automorphism group of G contains as many elements as G, then the automorphism group of G contains every element of G. Does this mean that every element of G gets mapped to itself if we define an automorphism φ: G to G?

If this is correct, then when Z(G) is trivial and we have Aut(G) ≈ Inn(G), then we know that the Inner automorphism group of G contains every element of G (every element of G commutes with every element of G) and also that the Automorphism group of G contains every element of G (Every element of G gets mapped to itself in the homomorphism φ: G to G).

Can someone let me know if I am understanding this right. Thanks!