Show that $M_t$ is a Standard Brownian Motion Let $M=(M_t)_{t\geq0}$ with $$M_t=\int_0^{\log\sqrt{1+2t}}e^s\text{d}B_s$$
where $(B_t)_{t\geq0}$ is a Standard Brownian Motion.
Show that $M$ is also a Standard Brownian Motion and compute $\mathbb{E}[\int_0^tM^6_s\text{d}M_s]$ and $\mathbb{E}[\left(\int_0^tM_s\text{d}M_s\right)^3]$.
Can we use Levy's Characterization Theorem?
 A: From the exchanges in the comments, one reaches the conclusion that $M_t=K_{A(t)}$ with 
$$
K_t=\int_0^t\mathrm e^s\mathrm dB_s,\qquad
A(t)=\log\sqrt{1+2t}.
$$ 
Since $(K_t)_{t\geqslant0}$ is a martingale and the (deterministic) function $A$ is nondecreasing, this is enough to show that $(M_t)_{t\geqslant0}$ is a martingale. Furthermore, for every $0\leqslant s\leqslant t$, by Itô's isometry,
the identity $$
(M_t-M_s)^2=\left(\int_{A(s)}^{A(t)}\mathrm e^u\mathrm dB_u\right)^2
$$
implies that
$$
\mathbb E((M_t-M_s)^2)=\mathbb E\left(\int_{A(s)}^{A(t)}\mathrm e^{2u}\mathrm du\right)=\frac{\mathrm e^{2A(t)}-\mathrm e^{2A(s)}}2.
$$
The function $A$ is tuned such that $\mathbb E((M_t-M_s)^2)=t-s$ hence $(M_t)_{t\geqslant0}$ is a Brownian motion. More generally, for every nonzero function $a$, $(M^a_t)_{t\geqslant0}$ is a Brownian motion, where
$$
M^a_t=\int_0^{A^{-1}(t)}a(s)\mathrm dB_s,\qquad A(t)=\int_0^{t}a(s)^2\mathrm ds.
$$
The two expectatons to be computed are direct since one can replace $(M_t)_{t\geqslant0}$ by  $(B_t)_{t\geqslant0}$ without changing the result. For example,
$$
X_t=\int_0^tB_t^6\mathrm dB_t
$$
is an odd functional of $(B_s)_{0\leqslant s\leqslant t}$, hence $\mathbb E(X_t)=0$. Likewise,
$$
Y_t=\int_0^tB_s\mathrm dB_s
$$
is a centered gaussian random variable hence $\mathbb E(Y_t^3)=0$.
On time-changed Brownian motions, see this.
Edit: Since $A$ is invertible and continuous, the sigma-algebras $\mathcal F_t^B=\sigma(B_s;s\leqslant t)$ and $\mathcal F_t^M=\sigma(M_s;s\leqslant t)$ are such that $\mathcal F^M_t=\mathcal F^B_{A(t)}$. For every $s\leqslant t$, $M_t=M_s+\Delta$ where
$$
\Delta=\int_{A(s)}^{A(t)}\mathrm e^u\mathrm dB_u.
$$
The increments of $(B_u)_{u\geqslant A(s)}$ are independent of $\mathcal F^B_{A(s)}$ hence $\mathbb E(\Delta\mid\mathcal F^B_{A(s)})=0$, which shows that $\mathbb E(M_t\mid\mathcal F^M_s)=M_s$. Hence $(M_t)_{t\geqslant0}$ is a martingale.
