let me work out the comments a bit further, starting from the identity (theorem 4 from <A HREF="http://statweb.stanford.edu/~cgates/PERSI/papers/random_matrices.pdf">Diaconis and Shahshahani</A>)

$$\int_{{\rm O}(n)}{\rm tr}\,g^p\,dg=\begin{cases}
0&{\rm if}\;\;p\;\;\text{is an odd integer}\\
1&{\rm if}\;\;p\;\;\text{is an even integer}
\end{cases}
$$

so the second integral of the OP evaluates to
$$\int_{{\rm O}(n)}g^p\,dg=\begin{cases}
0&{\rm if}\;\;p\;\;\text{is an odd integer}\\
n^{-1}\mathbb{1}&{\rm if}\;\;p\;\;\text{is an even integer}
\end{cases}
$$

Taylor expansion of the exponent in the first integral of the OP and term-by-term integration gives
$$\int_{{\rm O}(n)}e^g\,dg=n^{-1}\mathbb{1}\sum_{p=0}^\infty \frac{1}{(2p)!}=\frac{\cosh 1}{n}\mathbb{1}
$$

now for the third integral of the OP, we need the fourth-order tensor
$$\int_{{\rm O}(n)}(g^p)_{ij}(g^q)_{kl}\,dg=a_{pq}(n)\delta_{ij}\delta_{kl}+b_{pq}(n)\delta_{ik}\delta_{jl}+c_{pq}(n)\delta_{il}\delta_{jk}$$
so that the required integral takes the form
$$\int_{{\rm O}(n)}g^p Ag^q\,dg=a_{pq}(n)A+b_{pq}(n)A^{\rm t}+c_{pq}(n)\mathbb{1}\,{\rm tr}\,A
$$
by taking traces I find, using again theorem 4 from <A HREF="http://statweb.stanford.edu/~cgates/PERSI/papers/random_matrices.pdf">Diaconis and Shahshahani</A>, that
$$na_{pq}(n)+nb_{pq}(n)+n^2 c_{pq}(n)=\int_{{\rm O}(n)}{\rm tr}\,g^{p+q}\,dg=\begin{cases}
0&{\rm if}\;\;p+q\;\;\text{is an odd integer}\\
1&{\rm if}\;\;p+q\;\;\text{is an even integer}
\end{cases}
$$
$$n^2a_{pq}(n)+nb_{pq}(n)+nc_{pq}(n)=\int_{{\rm O}(n)}({\rm tr}\,g^p)({\rm tr}\,g^q)\,dg=\begin{cases}
p&{\rm if}\;\;p=q\;\;{\rm odd}\\
p+1&{\rm if}\;\;p=q\;\;{\rm even}\\
1&{\rm if}\;\;p\neq q\;\;\text{both even}\\
0&{\rm otherwise}
\end{cases}
$$
$$na_{pq}(n)+n^2 b_{pq}(n)+nc_{pq}(n)=\int_{{\rm O}(n)}{\rm tr}\,g^{|p-q|}\,dg=\begin{cases}
0&{\rm if}\;\;p+q\;\;\text{is an odd integer}\\
1&{\rm if}\;\;p+q\;\;\text{is an even integer}
\end{cases}
$$
Three equations with three unknowns give
$$
a_{pq}=\frac{-2}{(n-1)(n+2)},\;\;b_{pq}=c_{pq}=\frac{n}{(n-1)(n+2)}\;\;\mbox{if $p\neq q$ both odd}$$
$$
a_{pq}=b_{pq}=c_{pq}=\frac{1}{n+2},\;\;\mbox{if $p\neq q$ both even}$$
$$a_{pq}=1-(n+1)b_{pq},\;\;b_{pq}=c_{pq}=\frac{n-p}{(n-1)(n+2)}\;\;\mbox{if $p=q$ odd}$$
$$a_{pq}=1-(n+1)b_{pq},\;\;b_{pq}=c_{pq}=\frac{n-p-1}{(n-1)(n+2)}\;\;\mbox{if $p=q$ even}$$

That should solve the third integral, but I may well have made some error, so the OP will want to check the algebra.

---

> so my conclusion for the answer to the OP's question "when is $T_{nm}$
> conjugate to $T_{n'm'}$" is: whenever $n\neq m$ and $n'\neq m'$ and
> $n+m+n'+m'$ even