= Solution
Write $u(Y_i;\theta)=\partial_\theta\log f(Y_i;\theta)$ and $U_n(\theta)=\sum_i u(Y_i;\theta)$ for the individual and total <score functions>. Under the usual regularity assumptions, $\mathbb E_{\theta_0}u=0$ and $\operatorname{Cov}_{\theta_0}(u)=I(\theta_0)$, the nonsingular per-observation <Fisher information matrix>. The <central limit theorem> and <weak law of large numbers> give
$$
\frac{U_n(\theta_0)}{\sqrt n}\xrightarrow{d}N_d(0,I(\theta_0)),\qquad \frac{j_n(\theta_0)}n\xrightarrow{p}I(\theta_0).
$$
Rearranging the stipulated score expansion, including its remainder, gives
$$
\sqrt n(\widehat\theta_n-\theta_0)=\left(\frac{j_n(\theta_0)}n\right)^{-1}\frac{U_n(\theta_0)}{\sqrt n}+o_p(1).
$$
<Slutsky theorem> therefore proves the <asymptotic normality of a maximum likelihood estimator>:
$$
\boxed{\sqrt n(\widehat\theta_n-\theta_0)\xrightarrow{d}N_d(0,I(\theta_0)^{-1}).}
$$
A consistent plug-in <Fisher information matrix> yields the <Wald statistic> $W_n=n(\widehat\theta_n-\theta_0)^TI(\widehat\theta_n)(\widehat\theta_n-\theta_0)$. Under the <null hypothesis>, $W_n\xrightarrow{d}\chi_d^2$, so a test of asymptotic size $\alpha$ rejects when $W_n$ exceeds the $(1-\alpha)$ <quantile> of that <chi-squared distribution>. A consistent <Observed Fisher information> matrix gives the same limiting test.
With an $r$-dimensional parameter of interest $\psi$ and <nuisance parameter> $\lambda$, partition $I$ accordingly. The marginal asymptotic <covariance matrix> of $\widehat\psi$ is the $\psi\psi$ block of $I^{-1}$, whose inverse is the <efficient information>
$$
I_{\rm eff}=I_{\psi\psi}-I_{\psi\lambda}I_{\lambda\lambda}^{-1}I_{\lambda\psi}.
$$
Thus the nuisance-adjusted <Wald statistic> is $W_{\psi,n}=n(\widehat\psi-\psi_0)^TI_{\rm eff}(\widehat\theta)(\widehat\psi-\psi_0)$, with limiting null distribution $\chi_r^2$. Using $I_{\psi\psi}$ alone is generally incorrect when the <nuisance parameter> is estimated; it is valid when the cross-information vanishes.
For the <Inverse Gaussian distribution> in this problem, expand the exponent to obtain the log <likelihood>, up to terms independent of both parameters,
$$
\ell_n(\psi,\lambda)=\frac n2\log\psi-\frac\psi2\left(\frac{\sum_iY_i}{\lambda^2}-\frac{2n}{\lambda}+\sum_iY_i^{-1}\right).
$$
Its <score functions> are
$$
U_\lambda=\frac\psi{\lambda^3}(\sum_iY_i-n\lambda),\qquad U_\psi=\frac n{2\psi}-\frac12\left(\frac{\sum_iY_i}{\lambda^2}-\frac{2n}{\lambda}+\sum_iY_i^{-1}\right).
$$
For each positive $\psi$, the first score changes sign from positive to negative at $\lambda=\bar Y$, so this is the global maximizing <mean parameter of an exponential family>. Substituting it in the second score gives
$$
\boxed{\widehat\lambda=\bar Y,\qquad\widehat\psi_U=\left(\overline{Y^{-1}}-\frac1{\bar Y}\right)^{-1}.}
$$
The reciprocal function is <strictly convex>, so the denominator is positive for a nonconstant sample. The profile log <likelihood> is <strictly concave> in $\psi$ and tends to minus infinity at both endpoints, proving this is a maximum. Under continuous model sampling, a sample of size at least two is nonconstant <almost surely>. For a constant sample, including a one-observation sample, no finite joint <maximum-likelihood estimator> exists: the <likelihood> increases without bound as $\psi\to\infty$ at $\lambda=\bar Y$.
For one observation, differentiating again and using $\mathbb EY=\lambda$ gives
$$
I_{\psi\psi}=\frac1{2\psi^2},\qquad I_{\psi\lambda}=\mathbb E\left(\frac1{\lambda^2}-\frac Y{\lambda^3}\right)=0,\qquad I_{\lambda\lambda}=\psi\mathbb E\left(\frac{3Y}{\lambda^4}-\frac2{\lambda^3}\right)=\frac\psi{\lambda^3}.
$$
This <parameter orthogonality> proves that estimating $\lambda$ does not reduce the <efficient information> for $\psi$. The <Wald statistic> has the same algebraic form in the two cases:
$$
\boxed{W_\psi=\frac{n(\widehat\psi-\psi_0)^2}{2\widehat\psi^2}\xrightarrow[H_0]{d}\chi_1^2.}
$$
Using information evaluated at $\psi_0$ instead is another first-order equivalent convention.
\b[The printed claim of coincidence needs a first-order interpretation.] Known and unknown $\lambda$ need not give numerically identical <Wald statistics>, because their <maximum-likelihood estimators> differ. With $\lambda$ known, direct maximization gives
$$
\widehat\psi_K=\left(\overline{Y^{-1}}-\frac2\lambda+\frac{\bar Y}{\lambda^2}\right)^{-1},\qquad \widehat\psi_K^{-1}-\widehat\psi_U^{-1}=\frac{(\bar Y-\lambda)^2}{\lambda^2\bar Y}.
$$
For example, observations $1,2$ and known $\lambda=1$ give $\widehat\psi_K=4$ and $\widehat\psi_U=12$. At $\psi_0=1$, the displayed Wald convention gives $9/16$ and $121/144$, respectively. Thus exact equality is false. Under the <null hypothesis>, however, $\bar Y-\lambda=O_p(n^{-1/2})$, hence $\widehat\psi_K-\widehat\psi_U=O_p(n^{-1})$. Their centered, standardized estimates differ by $o_p(1)$, so their <Wald statistics> differ by $o_p(1)$. This is the valid asymptotic coincidence supplied by <inverse Gaussian shape estimation and nuisance orthogonality>.
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